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?

The most recent issue of The American Mathematical Monthly (August-September 2011, pp. 641-643) has a new proof by Luigi Pace based on elementary probability.  Here's the argument.
Let $X_1$ and $X_2$ be independent, identically distributed standard half-Cauchy random variables.  Thus their common pdf is $p(x) = \frac{2}{\pi (1+x^2)}$ for $x > 0$.
Let $Y = X_1/X_2$.  Then the pdf of $Y$ is, for $y > 0$, $$p_Y(y) = \int_0^{\infty} x p_{X_1} (xy) p_{X_2}(x) dx = \frac{4}{\pi^2} \int_0^\infty \frac{x}{(1+x^2 y^2)(1+x^2)}dx$$
$$=\frac{2}{\pi^2 (y^2-1)} \left[\log \left( \frac{1+x^2 y^2}{1+x^2}\right) \right]_{x=0}^{\infty} = \frac{2}{\pi^2} \frac{\log(y^2)}{y^2-1} = \frac{4}{\pi^2} \frac{\log(y)}{y^2-1}.$$
Since $X_1$ and $X_2$ are equally likely to be the larger of the two, we have $P(Y < 1) = 1/2$.  Thus 
$$\frac{1}{2} = \int_0^1 \frac{4}{\pi^2} \frac{\log(y)}{y^2-1} dy.$$  This is equivalent to $$\frac{\pi^2}{8} = \int_0^1 \frac{-\log(y)}{1-y^2} dy = -\int_0^1 \log(y) (1+y^2+y^4 + \cdots)  dy = \sum_{k=0}^\infty \frac{1}{(2k+1)^2},$$
which, as others have pointed out, implies $\zeta(2) = \pi^2/6$.  
