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.



Partitioning the set of vertices of a convex n-gon into nonintersecting polygons
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](https://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences) acting on the monomial $x^{2k-2}$, which is the constant $(2k-2)!$.


