Combinatoric proof for $\sum_{k=0}^n{n\choose k}\left(-1\right)^k\left(n-k\right)^4 = 0$ ($n\geqslant5$) I'm trying to prove the following:
For every $n \ge 5$:
$$\sum_{k=0}^n{n\choose k}\left(-1\right)^k\left(n-k\right)^4 = 0$$
I've tried cancelling one $(n-k)$, and got this:
$$n\sum_{k=0}^{n-1}{n-1\choose k}\left(-1\right)^k\left(n-k\right)^3 = 0$$
I've also tried expressing the first formula as such:
$$\sum_{k=0}^na_kb_k$$
Where $a_k = {n \choose k}\left(-1\right)^k$ and $b_k = \left(n-k\right)^4 = \sum_{j=0}^4{4\choose j}n^j\left(-k\right)^{4-j}$
It's easy to see that $\sum_{k=0}^n a_k = \left(1-1\right)^n = 0$ by the binomial theorem.
But I'm lost as to why this work only for n>=5. What am I missing?

Consider the sum
$$\sum_{k=0}^n {n\choose k} (-1)^k (n-k)^q$$
where $q\ge 0.$ This is
$$q! [z^q] \sum_{k=0}^n {n\choose k} (-1)^k \exp((n-k)z).$$
We obtain
$$q! [z^q]  \exp(nz)
\sum_{k=0}^n {n\choose k} (-1)^k \exp(-kz)
\\ = q! [z^q] \exp(nz) (1-\exp(-z))^n
\\ = q! [z^q] (\exp(z)-1)^n.$$
Now since $(\exp(z)-1)^n = z^n + \cdots$ this is zero when $q\lt n$ and
we have the claim. Observe that the EGF here yields $n! \times {q\brace
n}.$
