The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?

Just as a curiosity, a one-line-real-analytic-proof I found by combining different ideas from this thread and this question:
$$\begin{eqnarray*}\zeta(2)&=&\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy\\&=&\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\frac{\pi^2}{6}.\end{eqnarray*}$$

Update. By collecting pieces, I have another nice proof. By Euler's acceleration method or just an iterated trick like my $(1)$ here we get:
$$ \zeta(2) = \sum_{n\geq 1}\frac{1}{n^2} = \color{red}{\sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}}\tag{A}$$
and the last series converges pretty fast. Then we may notice that the last series comes out from a squared arcsine. That just gives another proof of $ \zeta(2)=\frac{\pi^2}{6}$.

A proof of the identity 
$$\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi}{2}\sum_{k\geq 0}\frac{(-1)^k}{2k+1}=\frac{\pi}{2}\cdot\frac{\pi}{4}$$
is also hidden in tired's answer here. For short, the integral
$$ I=\int_{-\infty}^{\infty}e^y\left(\frac{e^y-1}{y^2}-\frac{1}{y}\right)\frac{1}{e^{2y}+1}\,dy $$
is clearly real, so the imaginary part of the sum of residues of the integrand function has to be zero.

Still another way (and a very efficient one) is to exploit the reflection formula for the trigamma function: $$\psi'(1-z)+\psi'(z)=\frac{\pi^2}{\sin^2(\pi z)}$$ immediately leads to: $$\frac{\pi^2}{2}=\psi'\left(\frac{1}{2}\right)=\sum_{n\geq 0}\frac{1}{\left(n+\frac{1}{2}\right)^2}=4\sum_{n\geq 0}\frac{1}{(2n+1)^2}=3\,\zeta(2).$$

2018 update. We may consider that $\mathcal{J}=\int_{0}^{+\infty}\frac{\arctan x}{1+x^2}\,dx = \left[\frac{1}{2}\arctan^2 x\right]_0^{+\infty}=\frac{\pi^2}{8}$.
On the other hand, by Feynman's trick or Fubini's theorem
$$ \mathcal{J}=\int_{0}^{+\infty}\int_{0}^{1}\frac{x}{(1+x^2)(1+a^2 x^2)}\,da\,dx = \int_{0}^{1}\frac{-\log a}{1-a^2}\,da $$
and since $\int_{0}^{1}-\log(x)x^n\,dx = \frac{1}{(n+1)^2}$, by expanding $\frac{1}{1-a^2}$ as a geometric series we have
$$ \frac{\pi^2}{8}=\mathcal{J}=\sum_{n\geq 0}\frac{1}{(2n+1)^2}. $$
