Proving finite sum identity I could use a hint to prove 
$$\sum_{k=0}^n(-1)^k {{n}\choose{k}}(\alpha+k)^n=(-1)^nn! $$
I needed that identity and found it in "Table of integrals, series, and products" by Gradshteyn et al. But just grabbing it from there is not very satisfying.
 A: The LHS is $$S_n=\sum_{k=0}^n(-1)^k {n\choose k}P(k)$$ where $P$ denotes the polynomial $$P(x)=(x+\alpha)^n$$ The key idea of the proof is that, as every polynomial of degree at most $n$, $P$ is a linear combination of the falling factorials, that is, the polynomials $$Q_i(x)=x(x-1)\cdots(x-i+1)$$ for $0\leqslant i\leqslant n$, say, $$P(x)=\sum_{i=0}^nc_iQ_i(x)$$ The rest of the proof relies on some small play on binomial coefficients.
Fix some $i$, then, for every $k<i$, $Q_i(k)=0$ and, for every $i\leqslant k\leqslant n$, $${n\choose k}Q_i(k)=\frac{n!}{k!(n-k)!}\frac{k!}{(k-i)!}=\frac{n!}{(n-i)!}\frac{(n-i)!}{(k-i)!(n-k)!}=Q_i(n){n-i\choose k-i}$$
Summing these identities on $k$ yields $$\sum_{k=0}^n(-1)^k {n\choose k}Q_i(k)=Q_i(n)\sum_{k=i}^n(-1)^k {n-i\choose k-i}=(-1)^iQ_i(n)\sum_{k=0}^{n-i}(-1)^k {n-i\choose k}$$ By the binomial theorem, the sum on the RHS is $(1-1)^{n-i}$, that is, $0$ for every $i<n$ and $1$ for $i=n$. Summing these identities on $i$, one gets $$S_n=(-1)^nQ_n(n)c_n$$ It remains to note that $Q_n(n)=n!$ and that $c_n=1$ (why?), to conclude.
A: For real $\beta$, define the function $\phi(\beta):=\sum\limits_{k=0}^{n}(-)^k{n\choose k}(\alpha+k)^n e^{(\alpha+k)\beta}$. So, we equivalently have $\phi(\beta)=\frac{d^n}{d\beta^n}(e^{\alpha\beta}(1-e^\beta)^n)$. Now, observe from the general Leibniz product rule that the only non-zero term for $\phi(0)$ is the term $n!e^{\alpha\beta}(-e^{\beta})^n$ (why?). We are done.
A: Seeing that
$$\sum_{k=0}^n (-1)^k {n\choose k} (\alpha+k)^n$$
is a polynomial in $\alpha$ of  degree $n$ we may extract coefficients
where $0\le q\le n$:
$$[\alpha^q] \sum_{k=0}^n (-1)^k {n\choose k} (\alpha+k)^n
= \sum_{k=0}^n (-1)^k {n\choose k} {n\choose q} k^{n-q}
\\ = {n\choose q} \sum_{k=0}^n (-1)^k {n\choose k}  k^{n-q}.$$
Continuing we find
$${n\choose q} \sum_{k=0}^n (-1)^k {n\choose k}
(n-q)! [z^{n-q}] \exp(kz)
\\ = \frac{n!}{q!} [z^{n-q}]
\sum_{k=0}^n (-1)^k {n\choose k} \exp(kz)
= \frac{n!}{q!} [z^{n-q}] (1-\exp(z))^n.$$
Now  since $1-\exp(z)  = -z  -  \cdots$ we  have that  $(1-\exp(z))^n$
starts with $(-1)^n  z^n + \cdots.$ Hence we get  zero when $n-q\lt n$
or $q\gt 0.$ So all coefficients of the polynomial except the constant
one vanish. For the latter we get with $q=0$ the value
$$\frac{n!}{0!} [z^n] ((-1)^n z^n + \cdots) = (-1)^n n!$$
thus proving the claim.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 0}^{n}\pars{-1}^{k} {n \choose k}\pars{\alpha + k}^{n} & =
\sum_{k = 0}^{n}\pars{-1}^{k} {n \choose k}
\braces{\rule{0pt}{5mm}\bracks{z^{n}}n!\expo{\pars{\alpha + k}z}}
\\[5mm] & =
n!\bracks{z^{n}}\expo{\alpha z}
\sum_{k = 0}^{n}{n \choose k}\pars{-\expo{z}}^{k} =
n!\bracks{z^{n}}\expo{\alpha z}\pars{1 - \expo{z}}^{n}
\\[5mm] & =
\pars{-1}^{n}\, n!\bracks{z^{0}}\expo{\alpha z}
\pars{\expo{z} - 1 \over z}^{n}\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\pars{-1}^{n}\, n!\,
\lim_{z \to 0}\bracks{\expo{\alpha z}\pars{\expo{z} - 1 \over z}^{n}} =
\bbx{\pars{-1}^{n}\, n!}
\end{align}
