$$\zeta(2n) = \frac{\pi^{2n}}{4^n-1}\space\cdot\space\frac{n}{(2n)!}\cdot\frac{d}{dx^{2n}}\bigg[\ln\big(\sec(x)\big)\bigg]_{x\space=\space0}$$
I found a link between the Riemann Zeta function and the Maclaurin series of $\ln(\sec(x))$, and generalized it to create the above formula. Although the formula is very similar to the one given on Wikipedia which has the Bernoulli Coefficients in it, from what I could find on the internet, computation of Bernoulli coefficients by hand is a rather tedious task, as it requires the computation of a limit at $x=0$ every time. But here, $\ln(\sec(x))$ avoids that. This might not be revolutionary, but if one were to compute the value of say, $\zeta(12)$, from scratch, I would definitely prefer to go with this method of utilising the Maclaurin series of $\ln(\sec(x))$.
$$\frac{d}{dx^{2n}}\bigg[\ln\big(\sec(x)\big)\bigg]_{x\space=\space0}$$
seems to correspond to A000182 Tangent (or "Zag") numbers in which case
$$\frac{d}{dx^{2n}}\bigg[\ln\big(\sec(x)\big)\bigg]_{x\space=\space0}=(-4)^n \left(4^n-1\right) \zeta(1-2 n)\ ,$$
and it is true that
$$\zeta(2 n)=\frac{\pi^{2 n}}{\left(4^n-1\right)} \frac{n}{(2 n)!}(-4)^n \left(4^n-1\right) \zeta (1-2 n)\ .$$