Proving finite sum identity

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.