Prove that $\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^m=...$

Question

Let $\binom{n}{k}$ denotes the number of subsets with $k$ elements in $n$-elements set. Prove that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^m=\begin{cases} 0, & \text{ if } 0\le m \le n-1; \\ 2^n \cdot n!, & \text{ if } m=n. \end{cases}$$

Answer 1

By the binomial theorem we have $$ \sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}x^{2n-2k}=\left(x^{2}-1\right)^{n} $$ so $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}x^{n-2k}=x^{-n}\left(x^{2}-1\right)^{n} $$ and if we take the derivative we get $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}x^{n-2k-1}\left(n-2k\right)=-nx^{-n-1}\left(x^{2}-1\right)^{n}+n2x^{-n+1}\left(x^{2}-1\right)^{n-1} \tag{1} $$ now if we take $x=1 $ we can easily see that the RHS is $0$. From $(1)$ we can also observe, if we multiply $x$ in each side $$ \sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}x^{n-2k}\left(n-2k\right)=-nx^{-n}\left(x^{2}-1\right)^{n}+n2x^{-n+2}\left(x^{2}-1\right)^{n-1}$$ and now we can iterate the method. So you need to differentiate $n $ times to get a non zero term, which is $2^{n}n! $ (try for yourself to prove the last identity).