In detail, let a field $F=\mathrm{GF}(q)$ where $q$ is a power of an odd prime and let $a$ be a primitive element in $F$. I want to show that $a^{(q-1)/2}=-1$.
So, first we verify that $\frac{q-1}{2}\in\mathbb{Z}$ since $q\geq3$ and $q$ is odd. Now, by Lagrange's theorem if $x$ is a generator of some group $G$ then $$\mathrm{order}(<x>)\vert\mathrm{order}(G)$$ But here $a$ is a primitive element of $F$ and $\mathrm{order}(F)=q$ so we will hit 1 when $a$ is raised to $q-1$. Thus, $a^{(q-1)/2}$ is a root of the polynomial $x^2-1$ in $F$. Now, I am stuck here.
Thanks to Lord Shark the Unknown's hint I was able to reach to the end of the solution.
First, I was at the point where $a^{(q-1)/2}$ is a root of the polynomial $x^2-1$. But, this polynomial has two roots, namely $1$ and $-1$. To prove the result we need to show that $a^{(q-1)/2}$ cannot be $1$, but only $-1$.
Now, $1\in F$ since $F$ is a field. But since $a$ is a primitive element of $F$ if we start computing $a,a^2,a^3,\dots$ we can't hit $1$ for powers smaller than $q-1$. So $a^{(q-1)/2}\neq1$ and $a^{(q-1)/2}=-1$ is the only possible solution.