TITLE: Proving Combinatorical Summation: $n!=\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n$
QUESTION [7 upvotes]: been stuck with this question for the last few hours, any help would be appreciated.
$$
{\large n! = \sum_{k = 0}^{n}\left(-1\right)^{k}{\,n\, \choose \,k\,}
\left(\,n - k\,\right)^{n}}
$$
what I did:
$\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n=n!\sum_{k=0}^n\frac{(-1)^k(n-k)^n}{k!(n-k)!}.$
So we are left to prove $\sum_{k=0}^n\frac{(-1)^k(n-k)^n}{k!(n-k)!}=1$. tried doing so using induction, or treating the sum as geometric sequence (which turned out poorly)
Suggestions?

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{n! = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{n}}$

\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{n}
&=n!\bracks{\overbrace{\color{#c00000}{%
{1 \over n!}\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{n}}}
^{\ds{{\rm S}\pars{n,n} = 1}}}
\end{align}
  The $\color{#c00000}{\ds{\mbox{above red expression}}}$ is the
  Stirling Number of the Second Kind
  $\ds{{\rm S}\pars{n,n}}$. See formula $\pars{10}$ in the just cited link.

$\ds{{\rm S}\pars{n,n}}$ is the number of ways of partitioning a set of $\ds{n}$ elements into $\ds{n}$ nonempty sets which is obviously $\large\tt\mbox{equal to one}$.

$$\color{#66f}{\large%
n! = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\pars{n - k}^{n}}
$$