Ways to prove Euler's formula for $\zeta(2n)$

I usually prove it in the following way. Since: $$ \frac{t}{e^t-1} = \sum_{n\geq 0}\frac{B_n}{n!}t^n \tag{1}$$ it follows that: $$ \coth z-\frac{1}{z} = \sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1}.\tag{2}$$ On the other hand, by taking the logarithmic derivative of the Weierstrass product for the $\sinh $ function it follows that: $$\begin{eqnarray*} \coth z -\frac{1}{z} &=& \sum_{n\geq 1}\frac{d}{dz}\log\left(1+\frac{z^2}{n^2 \pi^2}\right)\\&=&\sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}\\&=&\sum_{n\geq 1}\sum_{m\geq 1}\frac{2(-1)^{m-1}z^{2m-1}}{\pi^{2m}n^{2m}} \\&=&\sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}\tag{3}\end{eqnarray*}$$ and we have the claim by comparing the coefficients in the RHSs of $(2)$ and $(3)$.

The intermediate identities are often very useful, too.

In order to compute the values $\zeta(2n),$ for $n\in\mathbb{N}^+,$ use equation Riemann's functional equation for the Zeta function and the double angle sine formula to obtain

\begin{equation}\tag{1} \zeta(2n)=\frac{(2\pi)^{2n}\zeta(1-2n)}{2\Gamma(2n)\cos(n\pi)}=\frac{(-1)^n(2\pi)^{2n}}{2(2n-1)!}\zeta(1-2n). \end{equation}

Using the fact that, for $n\in\mathbb{N},$ \begin{equation}\tag{2} B_n=(-1)^{n+1}n\zeta(1-n). \end{equation}

We can use (2) to rewrite (1) as \begin{equation}\tag{3} \zeta(2n)=\frac{(-1)^n(2\pi)^{2n}}{2(2n-1)!}\zeta(1-2n)=\frac{(-1)^{n+1}(2\pi)^{2n}}{2(2n)!}B_{2n}. \end{equation}