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.
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$\begingroup$ In Mathematica, one can check for specific k: With[{k = 7}, {Sum[(-1)^r Binomial[2 k - 2, r] (5 k - 2 - r)^(2 k - 2), {r, 0, 2 k - 2}] , (2 k - 2)! }] $\endgroup$– Per AlexanderssonCommented Apr 1, 2022 at 5:51
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$\begingroup$ Along the same lines as mathoverflow.net/q/417106 $\endgroup$– Max AlekseyevCommented Apr 1, 2022 at 11:12
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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 $(2k-2)!$.
answered Apr 1, 2022 at 7:10
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$\begingroup$ @ Timothy Budd -This approach is really interesting and very quick. I think here you have to find 2k-2 nd backwards difference ( which I did in my research to derive general formula) and differentiate the function 2k-2 times then equal the coefficient of the highest power ot n . $\endgroup$ Commented Apr 2, 2022 at 1:13