Xrfgtjtk

The probabilistic way:

This is $P[N_nleqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$.

By the central limit theorem, $Y_n=frac1{sqrt{n}}(X_1+cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_nleqslant 0]to P[Zleqslant0]$.

Finally, $P[Zleqslant0]=frac12$ and $[N_nleqslant n]=[Y_nleqslant 0]$ hence $P[N_nleqslant n]tofrac12$, QED.

The analytical way, completing your try:

Hence, I know that what I need to do is to find $limlimits_{ntoinfty}I_n$, where
$$
I_n=frac{e^{-n}}{n!}int_{0}^n (n-t)^ne^tdt.$$

To begin with, let $u(t)=(1-t)e^t$, then $I_n=dfrac{e^{-n}n^n}{n!}nJ_n$ with
$$
J_n=int_{0}^1 u(t)^nmathrm dt.
$$
Now, $u(t)leqslantmathrm e^{-t^2/2}$ hence
$$
J_nleqslantint_0^1mathrm e^{-nt^2/2}mathrm dtleqslantint_0^inftymathrm e^{-nt^2/2}mathrm dt=sqrt{frac{pi}{2n}}.
$$
Likewise, the function $tmapsto u(t)mathrm e^{t^2/2}$ is decreasing on $tgeqslant0$ hence $u(t)geqslant c_nmathrm e^{-t^2/2}$ on $tleqslant1/n^{1/4}$, with $c_n=u(1/n^{1/4})mathrm e^{-1/(2sqrt{n})}$, hence
$$
J_ngeqslant c_nint_0^{1/n^{1/4}}mathrm e^{-nt^2/2}mathrm dt=frac{c_n}{sqrt{n}}int_0^{n^{1/4}}mathrm e^{-t^2/2}mathrm dt=frac{c_n}{sqrt{n}}sqrt{frac{pi}{2}}(1+o(1)).
$$
Since $c_nto1$, all this proves that $sqrt{n}J_ntosqrt{fracpi2}$. Stirling formula shows that the prefactor $frac{e^{-n}n^n}{n!}$ is equivalent to $frac1{sqrt{2pi n}}$. Regrouping everything, one sees that $I_nsimfrac1{sqrt{2pi n}}nsqrt{fracpi{2n}}=frac12$.

Moral:
The probabilistic way is shorter, easier, more illuminating, and more fun.

Caveat:
My advice in these matters is, clearly, horribly biased.

Integration by parts yields
$$
frac{1}{k!}int_x^infty e^{-t},t^k,mathrm{d}t=frac{1}{k!}x^ke^{-x}+frac{1}{(k-1)!}int_x^infty e^{-t},t^{k-1},mathrm{d}ttag{1}
$$
Iterating $(1)$ gives
$$
frac{1}{n!}int_x^infty e^{-t},t^n,mathrm{d}t=e^{-x}sum_{k=0}^nfrac{x^k}{k!}tag{2}
$$
Thus, we get
$$
e^{-n}sum_{k=0}^nfrac{n^k}{k!}=frac{1}{n!}int_n^infty e^{-t},t^n,mathrm{d}ttag{3}
$$
Now, I will reproduce part of the argument I give here, which develops a full asymptotic expansion. Additionally, I include some error estimates that were previously missing.
$$
begin{align}
int_n^infty e^{-t},t^n,mathrm{d}t
&=n^{n+1}e^{-n}int_0^infty e^{-ns},(s+1)^n,mathrm{d}s\
&=n^{n+1}e^{-n}int_0^infty e^{-n(s-log(1+s)},mathrm{d}s\
&=n^{n+1}e^{-n}int_0^infty e^{-nu^2/2},s',mathrm{d}utag{4}
end{align}
$$
where $t=n(s+1)$ and $u^2/2=s-log(1+s)$.

Note that $frac{ss'}{1+s}=u$; thus, when $sge1$, $s'le2u$. This leads to the bound
$$
begin{align}
int_{sge1} e^{-nu^2/2},s',mathrm{d}u
&leint_{3/4}^infty e^{-nu^2/2},2u,mathrm{d}u\
&=frac2ne^{-frac98n}tag{5}
end{align}
$$
$(5)$ also show that
$$
int_{sge1}e^{-nu^2/2},mathrm{d}ulefrac2ne^{-frac98n}tag{6}
$$

For $|s|<1$, we get
$$
u^2/2=s-log(1+s)=s^2/2-s^3/3+s^4/4-dotstag{7}
$$
We can invert the series to get $s'=1+frac23u+O(u^2)$. Therefore,
$$
begin{align}
int_0^infty e^{-nu^2/2},s',mathrm{d}u
&=int_{sin[0,1]} e^{-nu^2/2},s',mathrm{d}u+color{red}{int_{s>1} e^{-nu^2/2},s',mathrm{d}u}\
&=int_0^inftyleft(1+frac23uright)e^{-nu^2/2},mathrm{d}u-color{darkorange}{int_{s>1}left(1+frac23uright)e^{-nu^2/2},mathrm{d}u}\
&+int_0^infty e^{-nu^2/2},O(u^2),mathrm{d}u-color{darkorange}{int_{s>1} e^{-nu^2/2},O(u^2),mathrm{d}u}\
&+color{red}{int_{s>1} e^{-nu^2/2},s',mathrm{d}u}\
&=sqrt{frac{pi}{2n}}+frac2{3n}+Oleft(n^{-3/2}right)tag{8}
end{align}
$$
The red and orange integrals decrease exponentially by $(5)$ and $(6)$.

Plugging $(8)$ into $(4)$ yields
$$
int_n^infty e^{-t},t^n,mathrm{d}t=left(sqrt{frac{pi n}{2}}+frac23right),n^ne^{-n}+O(n^{n-1/2}e^{-n})tag{9}
$$
The argument above can be used to prove Stirling's approximation, which says that
$$
n!=sqrt{2pi n},n^ne^{-n}+O(n^{n-1/2}e^{-n})tag{10}
$$
Combining $(9)$ and $(10)$ yields
$$
begin{align}
e^{-n}sum_{k=0}^nfrac{n^k}{k!}
&=frac{1}{n!}int_n^infty e^{-t},t^n,mathrm{d}t\
&=frac12+frac{2/3}{sqrt{2pi n}}+O(n^{-1})tag{11}
end{align}
$$

If you'd like to see formal solution using calculus methods check this article http://www.emis.de/journals/AMAPN/vol15/voros.pdf

The sum is related to the partial exponential sum, and thus to the incomplete gamma function,
$$begin{eqnarray*}
e^{-n} sum_{k=0}^{n} frac{n^k}{k!}
&=& e^{-n} e_n(n) \
&=& frac{Gamma(n+1,n)}{Gamma(n+1)},
end{eqnarray*}$$
since $e_n(x) = sum_{k=0}^n x^k/k! = e^x Gamma(n+1,x)/Gamma(n+1)$.
But
$$begin{eqnarray*}
Gamma(n+1,n) &=& sqrt{2pi}, n^{n+1/2}e^{-n}left(frac{1}{2} + frac{1}{3}sqrt{frac{2}{npi}} + Oleft(frac{1}{n}right) right).
end{eqnarray*}$$
The first term in the asymptotic expansion for $Gamma(n+1,n)$ can be found by applying the saddle point method to
$$Gamma(n+1,n) = int_n^infty dt, t^n e^{-t}.$$
The higher order terms are in principle straightforward to compute.
Using Stirling's approximation, we find
$$e^{-n} sum_{k=0}^{n} frac{n^k}{k!}
= frac{1}{2}
+ frac{1}{3}sqrt{frac{2}{npi}}
+ Oleft(frac{1}{n}right).$$
Thus, the limit is $1/2$, as found by @sos440 and @robjohn.
This limit is a special case of DLMF 8.11.13.

I just noticed a comment that suggests this be done using high school level math.
If this is a standard exercise at your high school, maybe they covered the incomplete gamma function! ;-)

$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle #1 rightrangle}
newcommand{braces}[1]{leftlbrace #1 rightrbrace}
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}
newcommand{dd}{{rm d}}
newcommand{isdiv}{,left.rightvert,}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{imp}{Longrightarrow}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{pars}[1]{left( #1 right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{,#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert #1 rightvert}
newcommand{yy}{Longleftrightarrow}$
begin{align}&color{#00f}{
lim_{n to infty}bracks{expo{-n}sum_{k = 0}^{n}{n^{k} over k!}}}
\[3mm]&=lim_{n to infty}bracks{expo{-n}sum_{k = 0}^{n}
exppars{klnpars{n} - lnpars{k!}}}
\[3mm]&=
lim_{n to infty}braces{expo{-n}sum_{k = 0}^{n}
exppars{nlnpars{n} - lnpars{n!} - {1 over 2n}bracks{k - n}^{2}}}
\[3mm]&=
lim_{n to infty}braces{expo{-n},{n^{n} over n!}sum_{k = 0}^{n}
exppars{-{1 over 2n}bracks{k - n}^{2}}}
\[3mm]&=
lim_{n to infty}braces{{expo{-n}n^{n} over n!}int_{0}^{n}
exppars{-{1 over 2n}bracks{k - n}^{2}},dd k}
\[3mm]&=
lim_{n to infty}bracks{{expo{-n}n^{n} over n!}int_{-n}^{0}
exppars{-,{k^{2} over 2n}},dd k}
=
lim_{n to infty}bracks{{expo{-n}n^{n} over n!},root{2n}
int_{-root{n}/2}^{0}exppars{-k^{2}},dd k}
\[3mm]&=
lim_{n to infty}bracks{{root{2}n^{n + 1/2}expo{-n} over n!}
int_{-infty}^{0}exppars{-k^{2}},dd k}
=
lim_{n to infty}bracks{{root{2}n^{n + 1/2}expo{-n} over n!}
,{root{pi} over 2}}
\[3mm]&=
half,lim_{n to infty}bracks{{root{2pi}n^{n + 1/2}expo{-n} over n!}}
=color{#00f}{Largehalf}
end{align}

I'll give you two hints:

1) Poisson distributions;

2) Central limit theorem

I am not aware of any other technique to solve the problem, so any other answer is appreciated.

I do not know how much this will help you.

For a given $n$, the result is $dfrac{Gamma(n+1,n)}{n Gamma(n)}$ which has a limit equal to $dfrac12$ as $ntoinfty$.

On this page there is a nice collection of evidence.

I add another proof which also uses the Stirling formula.

$displaystyle e^{-n}sumlimits_{k=0}^nfrac{n^k}{k!} = e^{-n}sumlimits_{k=0}^nfrac{k^k (n-k)^{n-k}}{k!(n-k)!} hspace{4cm}$ e.g. here

$displaystyle = limlimits_{ntoinfty} e^{-n}sumlimits_{k=1}^{n-1}frac{e^k e^{n-k}}{sqrt{2pi k (1+mathcal{O}(1/k))}sqrt{2pi (n-k)(1+mathcal{O}(1/(n-k)))}} $

$displaystyle = limlimits_{ntoinfty} frac{1}{2pi}frac{1}{n}sumlimits_{k=1}^{n-1}frac{1}{sqrt{frac{k}{n}left(1-frac{k}{n}right)}} =frac{1}{2pi} intlimits_0^1frac{dx}{sqrt{x(1-x)}}=frac{Gamma(frac{1}{2})^2}{2pi~Gamma(1)} = frac{1}{2}$