A sum for stirling numbers Pi, e.

In this identity

$$1-e{}^{2} = \displaystyle \sum _{n=0}^{\infty } \frac{(-1)^n(\pi )^{2 n}} {(2 n)!}\sum _{k=0}^{2 n} (-1)^{k} S_2(2 n,1-k+2 n),$$

$S_2$ is a Stirling number of the second kind. Would this be classified as a combinatoric identity? I suspect that the finite sum can be simplified. Is there a known closed form?

When odd powers taken instead of $2n$, it is equal to zero.

I'm not sure how to prove it analytically. Help with proof, or references to similar identities would be helpful.

Let us restate the sum that is to be evaluated: $$\sum_{n\ge 0} \frac{(-1)^n \pi^{2n}}{(2n)!} \sum_{k=0}^{2n} (-1)^k {2n \brace 2n+1-k}.$$

Recall the bivariate generating function of the Stirling numbers of the second kind: $$G(z, u) = \exp(u(\exp(z)-1)).$$

Substituting this into the sum yields $$\sum_{n\ge 0} \frac{(-1)^n \pi^{2n}}{(2n)!} \sum_{k=0}^{2n} (-1)^k (2n)! [z^{2n}] \frac{(\exp(z)-1)^{2n+1-k}}{(2n+1-k)!}$$ which is $$\sum_{n\ge 0} \frac{(-1)^n \pi^{2n}}{(2n)!} (2n)! [z^{2n}] \sum_{k=1}^{2n+1} (-1)^{2n+1-k} \frac{(\exp(z)-1)^k}{k!}.$$ We now extend the summation to the re-indexed $k=0$ which only contributes to $[z^0]$ by a value of $-1$. We cancel this by adding $1$ at the front to get $$1+\sum_{n\ge 0} \frac{(-1)^n \pi^{2n}}{(2n)!} (2n)! [z^{2n}] \sum_{k=0}^{2n+1} (-1)^{2n+1-k} \frac{(\exp(z)-1)^k}{k!}.$$

The inner exponential term starts at $z$ so we may re-write this as $$1 + \sum_{n\ge 0} \frac{(-1)^n \pi^{2n}}{(2n)!} (-1)^{2n+1} (2n)! [z^{2n}] \sum_{k=0}^\infty (-1)^k \frac{(\exp(z)-1)^k}{k!}$$ which is $$1 - \sum_{n\ge 0} \frac{(-1)^n \pi^{2n}}{(2n)!} (2n)! [z^{2n}] \sum_{k=0}^\infty (-1)^k \frac{(\exp(z)-1)^k}{k!}$$ and we finally obtain $$1 - \sum_{n\ge 0} \frac{(-1)^n \pi^{2n}}{(2n)!} (2n)! [z^{2n}] \exp(1-\exp(z)).$$

What we have here is an annihilated coefficient extractor that extracts the even terms from the series of $$\exp(1-\exp(z))$$ and sums and evaluates them at $z=i\pi.$ But the even terms are $$\frac{1}{2} \exp(1-\exp(z)) + \frac{1}{2} \exp(1-\exp(-z)).$$ This yields $$1 -\left(\frac{1}{2} \exp(1-\exp(i\pi)) + \frac{1}{2} \exp(1-\exp(-i\pi))\right)$$ or $$1 -\left(\frac{1}{2} \times \exp(2) + \frac{1}{2} \times \exp(2)\right) = 1-e^2.$$

The technique of annihilated coefficient extractors (ACE) is also employed at this MSE link I and this MSE link II.