What's the generalization for $\zeta(2n)$?

In this answer, it is shown that $$ \zeta(2n)=\frac{(-1)^{n-1}\pi^{2n}}{(2n+1)!}n+\sum_{k=1}^{n-1}\!\frac{(-1)^{k-1}\pi^{2k}}{(2k+1)!}\zeta(2n-2k) $$ which can be used recursively to compute $\zeta(2n)$ for all $n\ge1$.

This can easily be modified to $$ \frac{\zeta(2n)}{\pi^{2n}}=\frac{(-1)^{n-1}n}{(2n+1)!}+\sum_{k=1}^{n-1}\!\frac{(-1)^{k-1}}{(2k+1)!}\frac{\zeta(2n-2k)}{\pi^{2n-2k}} $$ which shows that $\frac{\zeta(2n)}{\pi^{2n}}\in\mathbb{Q}$.