TITLE: Compute $\sum \limits_{k=0}^{n}(-1)^{k}k^{m}\binom{n}{k}$ using Lagrange interpolation.
QUESTION [7 upvotes]: Using Lagrange interpolation (I think identity $\sum \limits_{k=0}^{n}k^{m}\prod \limits_{\substack{i=0\\i\neq k}}^{n}\frac{x-i}{k-i}=x^m$) shows that 
$$\sum \limits_{k=0}^{n}(-1)^{k}k^{m}\binom{n}{k}=0 \text{ if }\ 0≤m<n$$
and $$\sum \limits_{k=0}^{n}(-1)^{k}k^{n}\binom{n}{k}=(-1)^{n}n!$$ (same sum with $m=n$).
EDIT: I found it quite straightforward to prove first equality (just put x=0 into identity), the second part is definitely harder.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k}\,k^{m}{n \choose k} & =
\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\bracks{%
m!\oint_{\verts{z} = 1}{\exp\pars{kz} \over z^{m + 1}}\,{\dd z \over 2\pi\ic}}
\\[5mm] & =
m!\oint_{\verts{z} = 1}{1 \over z^{m + 1}}
\sum_{k = 0}^{n}{n \choose k}\pars{-\expo{z}}^{k}\,{\dd z \over 2\pi\ic} =
m!\pars{-1}^{n}\oint_{\verts{z} = 1}{\pars{\expo{z} - 1}^{n} \over z^{m + 1}}
\,{\dd z \over 2\pi\ic}
\\[5mm] & =
m!\pars{-1}^{n}\oint_{\verts{z} = 1}{1 \over z^{m + 1}}
\bracks{n!\sum_{k = 0}^{\infty}{k \brace n}{z^{k} \over k!}}
\,{\dd z \over 2\pi\ic} =
\bbx{\ds{\pars{-1}^{n}\,n!}\,{m \brace n}}
\end{align}

$\ds{a \brace b}$ is a Stirling Number of the Second Kind.