Evaluate the limit $\lim\limits_{n\to\infty}{\frac{n!}{n^n}\bigg(\sum_{k=0}^n{\frac{n^{k}}{k!}}-\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}\bigg)}$ 
Evaluate the limit
  $$ \lim_{n\rightarrow\infty}{\frac{n!}{n^{n}}\left(\sum_{k=0}^{n}{\frac{n^{k}}{k!}}-\sum_{k=n+1}^{\infty}{\frac{n^{k}}{k!}}  \right)} $$

I use $$e^{n}=1+n+\frac{n^{2}}{2!}+\cdots+\frac{n^{n}}{n!}+\frac{1}{n!}\int_{0}^{n}{e^{x}(n-x)^{n}dx}$$
but I don't know how to evaluate
$$ \lim_{n\rightarrow\infty}{\frac{n!}{n^{n}}\left(e^{n}-2\frac{1}{n!}\int_{0}^{n}{e^{x}(n-x)^{n}dx}  \right)      }$$

In this answer, it is shown, using integration by parts, that
$$
\sum_{k=0}^n\frac{n^k}{k!}=\frac{e^n}{n!}\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t\tag{1}
$$
Subtracting both sides from $e^n$ gives
$$
\sum_{k=n+1}^\infty\frac{n^k}{k!}=\frac{e^n}{n!}\int_0^n e^{-t}\,t^n\,\mathrm{d}t\tag{2}
$$
Substtuting $t=n(s+1)$ and $u^2/2=s-\log(1+s)$ gives us
$$
\begin{align}
\Gamma(n+1)
&=\int_0^\infty t^n\,e^{-t}\,\mathrm{d}t\\
&=n^{n+1}e^{-n}\int_{-1}^\infty e^{-n(s-\log(1+s))}\,\mathrm{d}s\\
&=n^{n+1}e^{-n}\int_{-\infty}^\infty e^{-nu^2/2}\,s'\,\mathrm{d}u\tag{3}
\end{align}
$$
and
$$
\begin{align}
\Gamma(n+1,n)
&=\int_n^\infty t^n\,e^{-t}\,\mathrm{d}t\\
&=n^{n+1}e^{-n}\int_0^\infty e^{-n(s-\log(1+s))}\,\mathrm{d}s\\
&=n^{n+1}e^{-n}\int_0^\infty e^{-nu^2/2}\,s'\,\mathrm{d}u\tag{4}
\end{align}
$$
Computing the series for $s'$ in terms of $u$ gives
$$
s'=1+\frac23u+\frac1{12}u^2-\frac2{135}u^3+\frac1{864}u^4+\frac1{2835}u^5-\frac{139}{777600}u^6+O(u^7)\tag{5}
$$
In the integral for $\Gamma(n+1)$, the odd powers of $u$ in $(5)$ are cancelled and the even powers of $u$ are integrated over twice the domain as in the integral for $\Gamma(n+1,n)$. Thus,
$$
\begin{align}
2\Gamma(n+1,n)-\Gamma(n+1)
&=\int_n^\infty t^n\,e^{-t}\,\mathrm{d}t-\int_0^n t^n\,e^{-t}\,\mathrm{d}t\\
&=n^{n+1}e^{-n}\int_0^\infty e^{-nu^2/2}\,2\,\mathrm{odd}(s')\,\mathrm{d}u\\
&=n^{n+1}e^{-n}\left(\frac4{3n}-\frac8{135n^2}+\frac{16}{2835n^3}+O\left(\frac1{n^4}\right)\right)\\
&=n^ne^{-n}\left(\frac43-\frac8{135n}+\frac{16}{2835n^2}+O\left(\frac1{n^3}\right)\right)\tag{6}
\end{align}
$$
Therefore, combining $(1)$, $(2)$, and $(6)$, we get
$$
\begin{align}
\frac{n!}{n^n}\left(\sum_{k=0}^n\frac{n^k}{k!}-\sum_{k=n+1}^\infty\frac{n^k}{k!}\right)
&=\frac{e^n}{n^n}\left(\int_n^\infty e^{-t}\,t^n\,\mathrm{d}t-\int_0^n e^{-t}\,t^n\,\mathrm{d}t\right)\\
&=\frac43-\frac{8}{135n}+\frac{16}{2835n^2}+O\left(\frac1{n^3}\right)\tag{7}
\end{align}
$$
