By considering the partial fraction decomposition of a certain family of expressions, I have managed to deduce the identity:
$$\sum_{k=0}^{n} (-1)^k {2n\choose k} (n-k)^{2m} \equiv 0 $$
where $n > m$, and $m,n \in \mathbb{N}$.
This seems like a surprisingly strong statement. Can this be proven another way?
Edit: the expression above closely resembles the formula for the number of surjective functions from $\{1,...,m\} \to \{1,...,n\}$, which is:
$$\sum_{k=0}^{n} (-1)^k {n\choose k} (n-k)^{m}$$