The material presented at this link on Zeta function values at even integers proposes a method to compute these that is based on Euler's work. I would like to present a short proof for your consideration (the goal being to keep it as simple as possible) and I'd be grateful to get confirmation of its correctness/admissibility e.g. by giving a reference.
We seek to prove that $$\zeta(2n) = \sum_{m\ge1} \frac{1}{m^{2n}} = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n}$$ where $B_{2n}$ are Bernoulli numbers. Rewrite this as $$\zeta(2n) = - \frac{1}{2} (2\pi i)^{2n} \frac{B_{2n}}{(2n)!}.$$ Now using the method presented here we introduce $$ f(z) = \frac{1}{z^{2n}} \pi \cot(\pi z)$$ and compute the integral of $f(z)$ along a circle of radius $R$ in the complex plane, where $R$ goes to infinity. We certainly have $|\pi \cot(\pi z)|<2\pi$ for $z$ on the circle and $R$ large enough. The term $1/z^{2n}$ is $\theta(1/R^{2n})$ so that the integral along the circle is $\theta(1/R^{2n-1})$ and vanishes in the limit. Using the Cauchy Residue theorem we thus obtain $$ 2 \zeta(2n) + \operatorname{Res}_{z=0} f(z) = \lim_{R\to\infty} \frac{1}{2\pi i} \int_{|z|=R} f(z) dz = 0.$$ The conclusion is that $$ \zeta(2n) = - \frac{1}{2} \operatorname{Res}_{z=0} f(z)$$ For the exponential generating function of the Bernoulli numbers we have $$\sum_{m=0}^\infty B_m \frac{t^m}{m!} = \frac{t}{e^t-1}$$ so that $$ \sum_{m=0}^\infty B_{2m} \frac{t^{2m}}{(2m)!} = \frac{1}{2} \left( \frac{t}{e^t-1} - \frac{t}{e^{-t}-1}\right) = \frac{1}{2} t \frac{e^{t/2}+e^{-t/2}}{e^{t/2}-e^{-t/2}}.$$ Setting $t=2\pi i z$, we obtain $$ \sum_{m=0}^\infty B_{2m} (2\pi i)^{2m} \frac{z^{2m}}{(2m)!} = \pi i z \frac{e^{\pi i z}+e^{-\pi i z}}{e^{\pi i z}-e^{-\pi i z}} = \pi z \cot(\pi z).$$ Putting it all together, we find $$\zeta(2n) = - \frac{1}{2} [z^{-1}] f(z) = - \frac{1}{2} [z^{2n-1}] \pi \cot(\pi z) = - \frac{1}{2} [z^{2n}] \pi z \cot(\pi z) = - \frac{1}{2} (2\pi i)^{2n} \frac{B_{2n}}{(2n)!},$$ where $[z^q]$ is the coefficient extraction operator for power series. This concludes the proof.
