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.
