Closed form or limiting form of an expression involving binomial coefficients This question leads to an application of the inclusion‒exclusion principle leading to this sum:
$$
\sum_{k=0}^n (-1)^k \binom n k (n-k)^x = (-1)^n \sum_{k=0}^n (-1)^k \binom n k k^x
$$
$$
\text{e.g. } \quad 1\cdot 6^{10} - 6\cdot 5^{10} +15\cdot 4^{10} - 6\cdot3^{10} + 1\cdot2^{10}
$$
Is there either a closed form or some sort of limiting form as $n\to\infty\text{?}$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{x} =
\pars{-1}^{n}\sum_{k=0}^{n}\pars{-1}^{k}{n \choose k}k^{x}:\ {\Large ?}}$.

\begin{align}
&\bbox[10px,#ffd]{\ds{%
\pars{-1}^{n}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}k^{x}}} =
\pars{-1}^{n}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}
\bracks{x!\oint_{\verts{z} = 1}
{\expo{kz} \over z^{x + 1}}\,{\dd z \over 2\pi\ic}}
\\[5mm] = &\
\pars{-1}^{n}\, x!\oint_{\verts{z} = 1}{1 \over z^{x + 1}}
\sum_{k = 0}^{n}{n \choose k}\pars{-\expo{z}}^{k}\,{\dd z \over 2\pi\ic} =
\pars{-1}^{n}\, x!\oint_{\verts{z} = 1}{\pars{1 - \expo{z}}^{n} \over z^{x + 1}}
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
x!\oint_{\verts{z} = 1}{\pars{\expo{z} - 1}^{n} \over z^{x + 1}}
\,{\dd z \over 2\pi\ic}
\end{align}

With the identity
  $\ds{\pars{\expo{z} - 1}^{n} = n!\sum_{\ell = 0}^{\infty}{\ell \brace n}{z^{\ell} \over \ell!}}$ where $\ds{{\ell \brace n}}$ is a
  Stirling Number of the Second Kind:

\begin{align}
&\bbox[10px,#ffd]{\ds{%
\pars{-1}^{n}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}k^{x}}} =
x!\, n!\sum_{\ell = 0}^{\infty}{\ell \brace n}{1 \over \ell!}\
\overbrace{\oint_{\verts{z} = 1}{1 \over z^{x - \ell + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{\delta_{\ell x}}} =
\bbx{n!\, {x \brace n}}
\end{align}

Note that $\ds{{x \brace n}_{\ x\ <\ n} = 0}$

