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.

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.

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.

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.