Question
How do you derive the following result: $$\sum_{n=1}^\infty (-1)^n\frac{\zeta(2n)}{\pi^{2n}} = \frac{1}{1-e^2}$$
More context
It is a well known fact that: $$\zeta(2n) = a_n\pi^{2n}$$ Where I'll call $a_n$ the coefficient of the zeta function at even integers. By using the explicit formula for the zeta function at even integers, we get that: $$a_n = \frac{(-1)^{n+1}2^{2n-1}B_{2n}}{(2n)!}$$ So by using this definition, we see that the original series is: $$\sum_{n=1}^\infty (-1)^n\frac{\zeta(2n)}{\pi^{2n}} = \sum_{n=1}^\infty (-1)^n a_n$$
Now, some of you may wonder; what about the non-alternating case? i.e. what is: $$\sum_{n=1}^\infty a_n =\sum_{n=1}^\infty \frac{(-1)^{n+1}2^{2n-1}B_{2n}}{(2n)!}$$
It turns out that this is nearly identical to the taylor series for $\cot(x)$:
$$\cot(x) = \frac{1}{x}\sum_{n=0}^\infty \frac{(-1)^{n}(2x)^{2n}B_{2n}}{(2n)!}$$
Plugging in 1, we get:
\begin{align}\cot(1)&= \sum_{n=0}^\infty \frac{(-1)^{n}2^{2n}B_{2n}}{(2n)!} \\\\ &= 1 + \sum_{n=1}^\infty \frac{(-1)^{n}2^{2n}B_{2n}}{(2n)!}\end{align}
Then, subtracting $-1$, multiplying by $-1$ and then dividing by 2, we get:
$$\frac{1-\cot(1)}{2} = \sum_{n=1}^\infty \frac{(-1)^{n+1}2^{2n-1}B_{2n}}{(2n)!}$$
...which is the desired series for the non-alternating case. So back to the question; how do you derive the series in alternating case?