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 $$ 2 \zeta(2n) + \mathrm{Res}_{z=0} f(z) = \lim_{R\to\infty} \frac{1}{2\pi i} \int_{|z|=R} f(z) dz = 0.$$
The conclusion is that
$$ \zeta(2n) = - \frac{1}{2} \mathrm{Res}_{z=0} f(z)$$
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.
Addendum Jul 12 2018. As remarked upon in the comments the bound does not work with a circle of radius $R$ even if passing between the poles. The reader of this post is invited to observe that the correct mechanics is to use a rectangle
$$\Gamma =\pm (N+1/2) \pm i N$$
with $N$ a large positive integer. The vertical segments pass midway between the poles. We show that the integral vanishes on two of these, with the other two being done similarly. We use two lines $\Gamma_1$ which is $N+1/2\pm iN$ (right vertical) and $\Gamma_2$ which is $\pm (N+1/2)+iN$ (top horizontal). Now on $\Gamma_1$ we find with $|t|\le N$
$$|\pi\cot(\pi (N+1/2) + \pi it)| =\pi\left|\frac{e^{i\pi (N+1/2) - \pi t}+e^{-i\pi (N+1/2) + \pi t}} {e^{i\pi (N+1/2) - \pi t}-e^{-i\pi (N+1/2) + \pi t}}\right| \\ = \pi\left|\frac{i(-1)^N e^{- \pi t} - i(-1)^N e^{\pi t}} {i(-1)^N e^{- \pi t} + i(-1)^N e^{\pi t}}\right| = \pi|\tanh(\pi t)|.$$
Observe that with $t$ real $\tanh(\pi t)$ has no poles and its norm is bounded by one. We thus get by ML the bound
$$\lim_{N\rightarrow\infty} 2N\pi / (N+1/2)^{2n} = 0$$
which vanishes as required. Continuing with $\Gamma_2$ we get for the norm of the trigonometric term (here $|t|\le N+1/2$)
$$|\pi\cot(\pi t + \pi i N)| = \pi\left|\frac{e^{i\pi t - \pi N} + e^{-i\pi t + \pi N}} {e^{i\pi t - \pi N} - e^{-i\pi t + \pi N}}\right| \le \pi\left|\frac{e^{\pi N}+e^{-\pi N}}{e^{\pi N}-e^{-\pi N}}\right| =\pi|\coth(\pi N)|.$$
There aren't any poles here either and this term is bounded above by $\pi\coth(\pi)$ because $N>1.$ Applying ML again yields
$$\lim_{N\rightarrow\infty} (2N+1) \pi\coth(\pi)/N^{2n} = 0.$$
The contribution from $\Gamma_2$ vanishes as well. Hence we have justified the construction used above.
