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How to prove that for matrix $Ainmathbb{R}^{n,n}$ we have
$$det A = det
begin{vmatrix}
x & x& x& ...&x&x\
1-x& 1&1 & ...&1& 1\
0& 1-x& 1& ...&1&1\
0& 0&1-x& ...&1&1\
&&ldots&& \
0& 0& 0& ...&1-x&1\
end{vmatrix} =x^n
$$

Subtract every column other than the last by the last column. Then move the last column to the first (this gives you a factor of $(-1)^{n-1}$ in the determinant). The result is a lower triangular matrix whose main diagonal is $(x,-x,-x,ldots,-x)$. Thus the determinant of the original matrix is $(-1)^{n-1}x(-x)^{n-1}=x^n$.

Let the determinant of $Ainmathbb R^{ntimes n}$ be a polynomial in $x$ denoted as $P_n(x); P_1(x)=det[x]=x$. Expand along the first column,

$det A=P_n(x)=
begin{vmatrix}
x& x & x & ldots & x& x \
1-x & 1 & 1&ldots &1& 1 \
0 & 1-x & 1 & ldots &1& 1 \
0 & 0 & 1-x & ldots &1& 1 \
vdots&vdots&vdots&&vdots&vdots\
0 & 0 & 0 & ldots&1-x&1 notag
end{vmatrix}_{n}\=xbegin{vmatrix}
1& 1 &ldots&1 \
1-x & 1&ldots &1\
0 & 1-x& ldots &1\
0 & 0& ldots &1\
vdots&vdots&&vdots\
0 & 0& ldots&1-xnotag
end{vmatrix}_{n-1}+(x-1)begin{vmatrix}
x& x&ldots & x\
1-x & 1&ldots &1\
0 & 1-x& ldots &1\
0 & 0& ldots &1\
vdots&vdots&&vdots\
0 & 0 & ldots&1-xnotag
end{vmatrix}_{n-1}$

Note that the first determinant is just $frac {P_{n-1}(x)}x$ and the second one $P_{n-1}(x)$.

$implies P_n(x)=xcdotfrac{P_{n-1}(x)}x+(x-1)P_{n-1}(x)\ =xP_{n-1}(x)\ =x^2P_{n-2}(x)\ =x^3P_{n-3}(x)\ =x^{n-1}P_{n-(n-1)}(x)\ =x^n$