How to prove that $\sum_{k=0}^n (-1)^k k^p \binom{n}{k}$ is non-zero for p>1

Question

Let $p\in \mathbb{N}_0$ and consider $\sum_{k=0}^n (-1)^k k^p \binom{n}{k}$. For $p=0$, this clearly vanishes as the sum then equals $(1-1)^n$. For $p=1$ and $n$ even, this sum also vanishes as one can replace the $k$ by $n-k$ and in this case $(-1)^{n-k}=(-1)^k$ holds. My intuition is that for $p\geq 2$, these cancelation phenomena do not occur but I have no idea how to prove or disprove this. I would greatly appreciate any help.

Answer 1

Suppose you have $p$ balls and $n$ bins, all distinct.

How many ways can you distribute the balls such that every bin is not empty?

That would be $n^p$ minus the number of ways such that at least one bin is empty. This can be found with inclusion-exclusion. The answer to this is exactly the expression you write about (possibly off by a factor of $-1$)

Clearly, even without counting fully we know that there exist ways to distribute balls with every box not empty when $p\geq n$ so the answer will be nonzero.

When $p<n$ the answer is zero. It is impossible to avoid having at least one empty box when having fewer balls than boxes. Your condition was wrong. With $0\leq p<n$ your sum equals zero. It is with $p\geq n$ that the sum becomes nonzero.

Answer 2

We have with $p\ge 2$ that the desired quantity is

$$\sum_{k=0}^n (-1)^k k^p {n\choose k} = p! [z^p] \sum_{k=0}^n (-1)^k {n\choose k} \exp(kz) \\ = p! [z^p] (1-\exp(z))^n = (-1)^n p! [z^p] (\exp(z)-1)^n.$$

Now the coefficients of $\exp(z)-1$ are all non-zero positive numbers so the coefficient on $[z^p]$ is non-zero as well, proving the claim, but only for $p\ge n$ because $\exp(z)-1 = z+\cdots$ and hence $(\exp(z)-1)^n$ starts at $z^n.$ (We get zero when $p\lt n.$) In fact we have for the coefficient that it is

$$(-1)^n p! \sum_{q_1+q_2+\cdots+q_n=p, q_j\ge 1} \prod_{j=1}^n \frac{1}{q_j!}.$$

Here we see once more that we must have $p\ge n$ for it not to be zero because if not the constraint equation has no solutions.

Note of course that the coefficient is also given by

$$(-1)^n n! p! [z^p] \frac{(\exp(z)-1)^n}{n!} = (-1)^n n! {p\brace n},$$

thereby providing a proof by inspection.