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I would like to know if the following formula is well known and get some references for it.

I don't know yet how to prove it (and I am working on it), but I am quite sure of its validity, after having performed a few symbolic computations with Maple.

Given $n$ square matrices $A_1,ldots,A_n$ of size $m<n$ :

$$sum_{p=1}^n(-1)^psum_{1leqslant i_1<cdots<i_pleqslant n}det(A_{i_1}+cdots+A_{i_p})=0$$

For example, if $A,B,C$ are three $2times2$ matrices, then :

$$det(A+B+C)-left[det(A+B)+det(A+C)+det(B+C)right]+det(A)+det(B)+det(C)=0$$

Let me outline two other proofs. Let me first rename your $m$ and $n$ as $n$
and $r$, since I find it confusing when $n$ is not the size of the square
matrices involved. So you are claiming the following:

Theorem 1. Let $mathbb{K}$ be a commutative ring. Let $ninmathbb{N}$
and $rinmathbb{N}$ be such that $n<r$. Let $A_{1},A_{2},ldots,A_{r}$ be
$ntimes n$-matrices over $mathbb{K}$. Then,
begin{equation}
sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }detleft( sumlimits_{iin I}A_{i}right) =0.
end{equation}

Notice that I've snuck in one more little change into your formula: I've added
the addend for $I=varnothing$. This addend usually doesn't contribute much,
because $detleft( sumlimits_{iinvarnothing}A_{i}right) =detleft(
0_{ntimes n}right) $ is usually $0$... unless $n=0$, in which case it
contributes $detleft( 0_{0times0}right) =1$ (keep in mind that there is
only one $0times0$-matrix and its determinant is $1$), and the whole equality
fails if this addend is missing.

A first proof of Theorem 1 appears in (the solution to) Exercise 6.53 in my
Notes on the combinatorial fundamentals of algebra, version of 10 January
2019. (To obtain
Theorem 1 from this exercise, set $G=left{ 1,2,ldots,rright} $.) The
main idea of this proof is that Theorem 1 holds not only for determinants, but
also for each of the $n!$ products that make up the determinant (assuming that
you define the determinant of an $ntimes n$-matrix as a sum over the $n!$
permutations); this is proven by interchanging summation signs and exploiting
discrete "destructive interference" (i.e., the fact that if $G$ is a finite
set and $R$ is a subset of $G$, then $sumlimits_{substack{Isubseteq
G;\Rsubseteq I}}left( -1right) ^{leftvert Irightvert }=
begin{cases}
1, & text{if }R=G;\
0, & text{if }Rneq G
end{cases}
$).

Let me now sketch a second proof of Theorem 1, which shows that it isn't
really about determinants. It is about finite differences, in a slightly more
general context than they are usually studied.

Let $M$ be any $mathbb{K}$-module. The dual $mathbb{K}$-module $M^{vee
}=operatorname{Hom}_{mathbb{K}}left( M,mathbb{K}right) $ of
$M$ consists of all $mathbb{K}$-linear maps $Mrightarrowmathbb{K}$. Thus,
$M^{vee}$ is a $mathbb{K}$-submodule of the $mathbb{K}$-module
$mathbb{K}^{M}$ of all maps $Mrightarrowmathbb{K}$. The $mathbb{K}
$-module $mathbb{K}^{M}$ becomes a commutative $mathbb{K}$-algebra (we just
define multiplication to be pointwise, i.e., the product $fg$ of two maps
$f,g:Mrightarrowmathbb{K}$ sends each $min M$ to $fleft( mright)
gleft( mright) inmathbb{K}$).

For any $dinmathbb{N}$, we let $M^{vee d}$ be the $mathbb{K}$-linear span
of all elements of $mathbb{K}^{M}$ of the form $f_{1}f_{2}cdots f_{d}$ for
$f_{1},f_{2},ldots,f_{d}in M^{vee}$. (For $d=0$, the only such element is
the empty product $1$, so $M^{vee0}$ consists of the constant maps
$Mrightarrowmathbb{K}$. Notice also that $M^{vee1}=M^{vee}$.) The elements
of $M^{vee d}$ are called homogeneous polynomial functions of degree $d$ on
$M$. The underlying idea is that if $M$ is a free $mathbb{K}$-module with a
given basis, then the elements of $M^{vee d}$ are the maps $Mrightarrow
mathbb{K}$ that can be expressed as polynomials of the coordinate functions
with respect to this basis; but the $mathbb{K}$-module $M^{vee d}$ makes
perfect sense whether or not $M$ is free.

We also set $M^{vee d}=0$ (the zero $mathbb{K}$-submodule of $mathbb{K}
^{M}$) for $d<0$.

For each $d in mathbb{Z}$, we define a $mathbb{K}$-submodule
$M^{vee leq d}$ of $mathbb{K}^M$ by
begin{equation}
M^{vee leq d} = sumlimits_{i leq d} M^{vee i} .
end{equation}
The elements of $M^{vee leq d}$ are called (inhomogeneous) polynomial
functions of degree $leq d$ on $M$.
The submodules $M^{vee leq d}$ satisfy
begin{equation}
M^{vee leq d} M^{vee leq e} subseteq M^{vee leq left(d+eright)}
end{equation}
for any integers $d$ and $e$.

For any $xin M$, we define the $mathbb{K}$-linear map $S_{x}:mathbb{K}
^{M}rightarrowmathbb{K}^{M}$ by setting
begin{equation}
left( S_{x}fright) left( mright) =fleft( m+xright) qquadtext{for
each }min Mtext{ and }finmathbb{K}^{M}.
end{equation}
This map $S_{x}$ is called a shift operator. It is an endomorphism of the
$mathbb{K}$-algebra $mathbb{K}^{M}$ and preserves all the $mathbb{K}
$-submodules $M^{vee leq d}$ (for all $dinmathbb{Z}$).

Moreover, for any $xin M$, we define the $mathbb{K}$-linear map $Delta
_{x}:mathbb{K}^{M}rightarrowmathbb{K}^{M}$ by $Delta_{x}
=operatorname*{id}-S_{x}$. Hence,
begin{equation}
left( Delta_{x}fright) left( mright) =fleft( mright) -fleft(
m+xright) qquadtext{for each }min Mtext{ and }finmathbb{K}^{M}.
end{equation}
This map $Delta_{x}$ is called a difference operator. The following crucial
fact shows that it "decrements the degree" of a polynomial function, similarly
to how differentiation decrements the degree of a polynomial:

Lemma 2. Let $x in M$. Then,
$Delta_{x}M^{vee d}subseteq M^{vee leq left( d-1right)}$
for each $dinmathbb{Z}$.

[Let me sketch a proof of Lemma 2:

Lemma 2 clearly holds for $d < 0$ (since $M^{vee d} = 0$ if $d < 0$).
Hence, it remains to prove Lemma 2 for $d geq 0$.
We shall prove this by induction on $d$.
The induction base is the case $d = 0$, which is easy to
check (indeed, each $f in M^{vee 0}$ is a constant map, and thus
satisfies $Delta_x f = 0$; therefore,
$Delta_{x}M^{vee 0} = 0 subseteq M^{vee leq left( 0-1right) }$).

For the induction step, we fix some nonnegative integer $e$, and assume
that Lemma 2 holds for $d = e$. We must then show that Lemma 2
holds for $d = e+1$.

We have assumed that Lemma 2 holds for $d = e$.
In other words, we have
$Delta_{x}M^{vee e}subseteq M^{vee leq left( e-1right)}$.

Our goal is to show that Lemma 2
holds for $d = e+1$. In other words, our goal is to show
that
$Delta_{x}M^{vee left(e+1right)}subseteq M^{vee leq e}$.

But the $mathbb{K}$-module $M^{vee left(e+1right)}$ is
spanned by maps of the form $fg$ with $fin M^{vee e}$ and
$gin M^{vee}$ (since it is spanned by products of the
form $f_1 f_2 cdots f_{e+1}$ with
$f_1, f_2, ldots, f_{e+1} in M^{vee}$, but each such
product can be rewritten in the form $fg$
with $f = f_1 f_2 cdots f_e in M^{vee e}$ and
$g = f_{e+1} in M^{vee}$).
Hence, it suffices to show that
$Delta_x left( fg right) in M^{vee leq e}$
for each $fin M^{vee e}$ and
$gin M^{vee}$.

Let us first notice that if $g in M^{vee}$ is arbitrary,
then $Delta_x g$ is the constant map whose value is
$- gleft(xright)$
(because each $m in M$ satisfies
begin{equation}
left(Delta_x gright) left(mright)
= gleft(mright) - underbrace{gleft(m+xright)}_{substack{=gleft(mright) + gleft(xright)\ text{(since }g text{ is } mathbb{K}text{-linear)}}}
= gleft(mright) - left(gleft(mright) + gleft(xright)right)
= - gleft(xright)
end{equation}
), and thus belongs to $M^{vee 0}$.
In other words, $Delta_x M^{vee} subseteq M^{vee 0}$.

For each $f in mathbb{K}^M$ and $g in mathbb{K}^M$,
we have
begin{align*}
Delta_{x}left( fgright) & =left( operatorname*{id}-S_{x}right)
left( fgright) qquadleft( text{since }Delta_{x}=operatorname*{id}
-S_{x}right) \
& =fg-underbrace{S_{x}left( fgright) }_{substack{=left( S_{x}fright)
left( S_{x}gright) \text{(since }S_{x}text{ is an endomorphism}
\text{of the }mathbb{K}text{-algebra }mathbb{K}^{M}text{)}}}\
& =fg-left( S_{x}fright) left( S_{x}gright) =underbrace{left(
f-S_{x}fright) }_{=left( operatorname*{id}-S_{x}right) f}g+left(
S_{x}fright) underbrace{left( x-S_{x}gright) }_{=left(
operatorname*{id}-S_{x}right) g}\
& =left( underbrace{left( operatorname*{id}-S_{x}right) }_{=Delta
_{x}}fright) g+left( S_{x}fright) left( underbrace{left(
operatorname*{id}-S_{x}right) }_{=Delta_{x}}gright) \
& =left( Delta_{x}fright) g+left(
underbrace{S_{x}}_{substack{=operatorname*{id}-Delta_{x}\
text{(since }Delta
_{x}=operatorname*{id}-S_{x}text{)}}}fright) left( Delta_{x}gright)
\
& =left( Delta_{x}fright) g+underbrace{left( left(
operatorname*{id}-Delta_{x}right) fright) }_{=f-Delta_{x}f}left(
Delta_{x}gright) \
& =left( Delta_{x}fright) g+left( f-Delta_{x}fright) left(
Delta_{x}gright) \
& =left( Delta_{x}fright) g+fleft( Delta_{x}gright) -left(
Delta_{x}fright) left( Delta_{x}gright) .
end{align*}
Hence, for each $fin M^{vee e}$ and $gin M^{vee}$, we have
begin{align*}
Delta_{x}left( fgright) & =left( Delta_{x}underbrace{f}_{in
M^{vee e}}right) underbrace{g}_{in M^{vee}}+underbrace{f}_{in M^{vee
e}}left( Delta_{x}underbrace{g}_{in M^{vee}}right) -left( Delta
_{x}underbrace{f}_{in M^{vee e}}right)
left( Delta_{x}underbrace{g}_{in M^{vee}}right) \
& inunderbrace{left( Delta_{x}M^{vee e}right) }_{subseteq M^{vee
leqleft( e-1right) }}M^{vee}+M^{vee e}underbrace{left( Delta
_{x}M^{vee}right) }_{subseteq M^{vee0}}-underbrace{left( Delta
_{x}M^{vee e}right) }_{subseteq M^{veeleqleft( e-1right) }
}underbrace{left( Delta_{x}M^{vee}right) }_{subseteq M^{vee0}}\
& subsetequnderbrace{M^{veeleqleft( e-1right) }M^{vee}}_{subseteq
M^{veeleq e}}+underbrace{M^{vee e}M^{vee0}}_{subseteq M^{vee
e}subseteq M^{veeleq e}}-underbrace{M^{veeleqleft( e-1right)
}M^{vee0}}_{subseteq M^{veeleqleft( e-1right) }subseteq M^{veeleq
e}}\
& subseteq M^{veeleq e}+M^{veeleq e}-M^{veeleq e}subseteq M^{veeleq
e}.
end{align*}
This proves that $Delta_{x}left( M^{veeleft( e+1right) }right)
subseteq M^{veeleq e}$, as we intended to prove.

Thus, the induction step is complete, and Lemma 2 is proven.]

The following fact follows by induction using Lemma 2:

Corollary 3. Let $rinmathbb{N}$. Let $x_{1},x_{2},ldots,x_{r}$ be $r$
elements of $M$. Then,
begin{equation}
Delta_{x_{1}}Delta_{x_{2}}cdotsDelta_{x_{r}}M^{vee d}subseteq
M^{vee leq left( d-rright) }
end{equation}
for each $dinmathbb{Z}$.

And as a consequence of this, we obtain the following:

Corollary 4. Let $rinmathbb{N}$. Let $x_{1},x_{2},ldots,x_{r}$ be $r$
elements of $M$. Then,
begin{equation}
Delta_{x_{1}}Delta_{x_{2}}cdotsDelta_{x_{r}}M^{vee d}=0
end{equation}
for each $dinmathbb{Z}$ satisfying $d<r$.

[In fact, Corollary 4 follows immediately from Corollary 3, because $d<r$
implies $M^{vee leq left( d-rright) }=0$.]

To make use of Corollary 4, we want a more-or-less explicit expression for how
$Delta_{x_{1}}Delta_{x_{2}}cdotsDelta_{x_{r}}$ acts on maps in
$mathbb{K}^{M}$. This is the following fact:

Proposition 5. Let $rinmathbb{N}$. Let $x_{1},x_{2},ldots,x_{r}$ be $r$
elements of $M$. Then,
begin{equation}
left( Delta_{x_{1}}Delta_{x_{2}}cdotsDelta_{x_{r}}fright) left(
mright) =sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }fleft( m+sumlimits_{iin I}x_{i}right)
qquadtext{for each }min Mtext{ and }finmathbb{K}^{M}.
end{equation}

[Proposition 5 can be proven by induction over $r$, where the induction step
involves splitting the sum on the right hand side into the part with the $I$
that contain $r$ and the part with the $I$ that don't. But there is also a
slicker argument, which needs some preparation. The maps $S_{x}in
operatorname{End}_{mathbb{K}}left( mathbb{K}^{M}right) $ for
different elements $xin M$ commute; better yet, they satisfy the
multiplication rule $S_{x}S_{y}=S_{x+y}$ (as can be checked immediately).
Hence, by induction over $leftvert Irightvert $, we conclude that if $I$
is any finite set, and if $x_{i}$ is an element of $M$ for each $iin I$, then
begin{equation}
prodlimits_{iin I}S_{x_{i}}=S_{sumlimits_{iin I}x_{i}}
qquad text{in the ring } operatorname{End}_{mathbb{K}} left(mathbb{K}^Mright) .
end{equation}
I shall refer to this fact as the S-multiplication rule.

Now, let us prove Proposition 5. Let $x_{1},x_{2},ldots,x_{r}$ be $r$
elements of $M$. Recall the well-known formula
begin{equation}
prodlimits_{iinleft{ 1,2,ldots,rright} }left( 1-a_{i}right)
=sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }prodlimits_{iin I}a_{i},
end{equation}
which holds whenever $a_{1},a_{2},ldots,a_{r}$ are commuting elements of some
ring. Applying this formula to $a_{i}=S_{x_{i}}$, we obtain
begin{equation}
prodlimits_{iinleft{ 1,2,ldots,rright} }left( operatorname*{id}
-S_{x_{i}}right) =sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left(
-1right) ^{leftvert Irightvert }prodlimits_{iin I}S_{x_{i}}
end{equation}
(since $S_{x_{1}},S_{x_{2}},ldots,S_{x_{r}}$ are commuting elements of the
ring $operatorname{End}_{mathbb{K}}left( mathbb{K}^{M}right)
$). Thus,
begin{align*}
Delta_{x_{1}}Delta_{x_{2}}cdotsDelta_{x_{r}} & =prodlimits_{iinleft{
1,2,ldots,rright} }underbrace{Delta_{x_{i}}}
_{substack{=operatorname*{id}-S_{x_{i}}\text{(by the definition of }
Delta_{x_{i}}text{)}}}=prodlimits_{iinleft{ 1,2,ldots,rright} }left(
operatorname*{id}-S_{x_{i}}right) \
& =sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }underbrace{prodlimits_{iin I}S_{x_{i}}}
_{substack{=S_{sumlimits_{iin I}x_{i}}\text{(by the S-multiplication rule)}
}}=sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }S_{sumlimits_{iin I}x_{i}}.
end{align*}
Hence, for each $min M$ and $finmathbb{K}^{M}$, we obtain
begin{align*}
& left( Delta_{x_{1}}Delta_{x_{2}}cdotsDelta_{x_{r}}fright) left(
mright) \
& =left( sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }S_{sumlimits_{iin I}x_{i}}fright) left( mright)
\
& =sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }underbrace{left( S_{sumlimits_{iin I}x_{i}}fright)
left( mright) }_{substack{=fleft( m+sumlimits_{iin I}x_{i}right)
\text{(by the definition of the shift operators)}}}\
& =sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }fleft( m+sumlimits_{iin I}x_{i}right) .
end{align*}
Thus, Proposition 5 is proven.]

We can now combine Corollary 4 with Proposition 5 and obtain the following:

Corollary 6. Let $x_{1},x_{2},ldots,x_{r}$ be $r$ elements of $M$. Let
$dinmathbb{Z}$ be such that $d<r$. Let $fin M^{vee d}$ and $min M$. Then,
begin{equation}
sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }fleft( m+sumlimits_{iin I}x_{i}right) =0.
end{equation}

[Indeed, Corollary 6 follows from the computation
begin{align*}
& sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }fleft( m+sumlimits_{iin I}x_{i}right) \
& =underbrace{left( Delta_{x_{1}}Delta_{x_{2}}cdotsDelta_{x_{r}
}fright) }_{substack{=0\text{(by Corollary 4, since } f in M^{vee d} text{)}}}left( mright)
qquadleft( text{by Proposition 5}right) \
& =0.
end{align*}
]

Finally, let us prove Theorem 1. The matrix ring $mathbb{K}^{ntimes n}$ is a
$mathbb{K}$-module. Let $M$ be this $mathbb{K}$-module $mathbb{K}^{ntimes
n}$. For each $i,jinleft{ 1,2,ldots,nright} $, we let $x_{i,j}$ be the
map $Mrightarrowmathbb{K}$ that sends each matrix $M$ to its $left(
i,jright) $-th entry; this map $x_{i,j}$ is $mathbb{K}$-linear and thus
belongs to $M^{vee}$.

It is easy to see that the map $det:mathbb{K}^{ntimes n}rightarrow
mathbb{K}$ (sending each $ntimes n$-matrix to its determinant) is a
homogeneous polynomial function of degree $n$ on $M$; indeed, it can be
represented in the commutative $mathbb{K}$-algebra $mathbb{K}^M$ as
begin{equation}
det=sumlimits_{sigmain S_{n}}left( -1right) ^{sigma}x_{1,sigmaleft(
1right) }x_{2,sigmaleft( 2right) }cdots x_{n,sigmaleft( nright)
},
end{equation}
where $S_{n}$ is the $n$-th symmetric group, and where $left( -1right)
^{sigma}$ denotes the sign of a permutation $sigma$. In other words,
$detin M^{vee n}$. Hence, Corollary 6 (applied to $x_{i}=A_{i}$, $d=n$,
$f=det$ and $m=0$) yields
begin{equation}
sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }detleft( 0+sumlimits_{iin I}A_{i}right) =0.
end{equation}
In other words,
begin{equation}
sumlimits_{Isubseteqleft{ 1,2,ldots,rright} }left( -1right)
^{leftvert Irightvert }detleft( sumlimits_{iin I}A_{i}right) =0.
end{equation}
This proves Theorem 1. $blacksquare$

Given integers $n > m > 0$, let $[n]$ be a short hand for the set ${1,ldots,n}$.

For any $t in mathbb{R}$ and $x_1, ldots, x_n in mathbb{C}$, we have the identity

$$prod_{k=1}^n (1 - e^{tx_k}) = sum_{P subset [n]} (-1)^{|P|} e^{tsum_{kin P} x_k}$$

Treat both sides as function of $t$. Expand against $t$, one notice on LHS, coefficients in front of $t^k$ vanishes whenever $k < n$.
By comparing coefficients of $t^m$, we obtain:

$$ 0 = sum_{Psubset [n]} (-1)^{|P|} left(sum_{kin P} x_kright)^mtag{*1}$$

Notice RHS is a polynomial function in $x_1,ldots,x_n$ with integer coefficients. Since it evaluates to $0$ for all $(x_1,ldots,x_n) in mathbb{C}^n$, it is valid as a polynomial identity in $n$ indeterminates with integer coefficients. As a corollary, it is valid as an algebraic identity when $x_1, x_2, ldots, x_n$ are elements taken from any commutative algebra.

Let $V$ be a vector space over $mathbb{C}$ spanned by
elements $eta_1, ldots, eta_m$ and $bar{eta}_1,ldots,bar{eta}_m$.

Let $Lambda^{e}(V) = bigoplus_{k=0}^n Lambda^{2k}(V)$ be the 'even' portion
of its exterior algebra. $Lambda^{e}(V)$ itself is a commutative algebra.

For any $m times m$ matrix $A$, let $tilde{A} in Lambda^e(V)$ be the element defined by:

$$A = (a_{ij}) quadlongrightarrowquad tilde{A} = sum_{i=1}^msum_{j=1}^m a_{ij}bar{eta}_i wedge eta_j$$

Notice the $m$-fold power of $tilde{A}$ satisfies an interesting identity:

$$tilde{A}^m = underbrace{tilde{A} wedge cdots wedge tilde{A}}_{m text{ times}} = det(A) omega
quadtext{ where }quad
omega = m!, bar{eta}_1 wedge eta_1 wedge cdots wedge bar{eta}_m wedge eta_mtag{*2}$$

Given any $n$-tuple of matrices $A_1, ldots, A_n in M_{mtimes m}(mathbb{C})$, if we substitute $x_k$ in $(*1)$ by $tilde{A}_k$ and apply $(*2)$, we find

$$
sum_{Psubset [n]} (-1)^{|P|} left(sum_{kin P} tilde{A}_kright)^m
= sum_{Psubset [n]} (-1)^{|P|} detleft(sum_{kin P} A_kright)omega
= 0
$$
Extracting the coefficient in front of $omega$, the desired identity follows:
$$sum_{Psubset [n]} (-1)^{|P|} detleft(sum_{kin P} A_kright) = 0$$

HINT:

The determinant of an $ntimes n$ matrix is a form of degree $n$. Forms come from multilinear forms.

Consider $M$ an abelian group. For $a in M$, denote by $a^{[n]}$ the element $aotimes a otimes ldots otimes ain M^{otimes n}$. Let now $a_iin M$, $i in I$, finitely many elements in $M$. Let's try to find
$$sum_{Jsubset I}(-1)^{|I|-|J|}(sum_{i in J} a_i)^{[n]}$$

Consider a product $a_{i_1}otimes ldots otimes a_{i_n}$. It appears in the above sum with the coefficient
$$sum_{Jsubset K subset I}(-1)^{|I| - |J|}$$ where $J={i_1, ldots, i_n }$. This is $0$ for $Jne I$ and $1$ for $J=I$. ( a Möbius function)

Therefore
$$sum_{Jsubset I}(-1)^{|I|-|J|}(sum_{i in J} a_i)^{[n]}=sum_{phicolon {1,ldots n}to I,phi text{surjective}}a_{phi(1)}otimes ldots a_{phi(n)}$$

Particular cases:

1. $|I|>n$, we get $0$, the result desired.




2. $|I|=n$, we get $sum_{phicolon {1,ldots n}to I,phi text{bijective}}a_{phi(1)}otimes ldots a_{phi(n)}$