Combinatorics – Partitioning Vertices of a Convex n-Gon into Nonintersecting Polygons

Question

How can you prove that
$$
\sum_{r=0}^{2k-2} (-1)^r \binom{2k-2}{r} (5k-2-r)^{2k-2} =(2k-2)!
$$
This result I have obtained by comparing results of two different approaches for the partitioning of the set of vertices of a convex n-gon into nonintersecting polygons.

Answer 1

Note that more generally $$ \nabla^{2k-2}[x^{2k-2}](x) = \sum_{r=0}^{2k-2}(-1)^r\binom{2k-2}{r}(x-r)^{2k-2} $$ is the order-$(2k-2)$ backwards finite difference operator acting on the monomial $x^{2k-2}$, which is the constant $(2k-2)!$.