Consider a compact Riemann surface $X$ of genus $g$.
It is well-known that its fundamental group $\pi_1(X)$ is the free group  on the generators $a_1,b_1,...,a_g,b_g$ divided out by the normal subgroup generated by  the single relator $[a_1,b_1]\cdot \ldots\cdot [a_g,b_g]$.
(This has of course nothing to do with the complex structure of $X$, but may be  computed by considering the underlying topological manifold as a cell complex.)
This group is trivial for $g=0$ and free abelian on two generators for $g=1$. 

For $g\geq 2$, however, I had always taken for granted  that it is not free  but I have just realized that I cannot prove that. 

So, although I guess   the answer is no, I'll ask my official question in an open way : Is   $\pi_1(X)$ free for $g\geq 2$ ?  

Edit: Users have now brilliantly solved the problem in multiple ways.
Non-freeness is definitely established, with 12 proofs plus sketches of proofs in the comments!
It is clearly impossible to select in a reasonable way an answer for "acceptance" among all these great answers .
Since the software forces me to make only one choice, I have chosen Daniel's answer because of its merit, but also because it acknowledges Vitali's  contribution: Vitali was  the first to sketch a solution (in the comments) . 

As per Theo's request, I'm posting this as an answer, though it's largely an expansion on Vitali's comment.  Let $F_n$ be the free group on $n$ letters; $K(F_n, 1)$ is a wedge of $n$ circles and so has vanishing cohomology in degrees $>1$.  On the other hand, if $X$ is a compact Riemann surface of genus $g>1$, $X$ is a $K(\pi_1(X), 1)$ as its universal cover is the upper-half plane, which is contractible.  But then $$H^2(\pi_1(X), \mathbb{Z})=H^2_{sing}(X, \mathbb{Z})=\mathbb{Z},$$ which is non-zero.  In particular, $\pi_1(X)$ has non-vanishing cohomology in degree $2$ and so is not free, as Vitali says.