In a finite field product of non-square elements is a square I came across one problem in a finite field as follows:

Let $F$ be a finite field. Show that if $a, b\in F$ both are non-squares, then $ab$ is a square. 

I wanted to prove it by using the idea of Biquadratic field extension. But there is no biquadratic extension over finite fields. Please, any hints for proving above fact? Thanks. 

Zero is a square, hence we may assume that both $a$ and $b$ belong to $\mathbb{F}^*$.
$\mathbb{F}^*$ is a cyclic group with order $q-1$: if the characteristic is $2$, every element of $\mathbb{F}$ is a square and there is nothing to prove. Otherwise, we may assume that $\mathbb{F}^*$ is generated by some element $g$, and in such a case the quadratic residues in $\mathbb{F}^*$ are the elements of the form $g^{\text{even}}$ and the non-quadratic residues are the elements of the form $g^{\text{odd}}$. Since $\text{odd}+\text{odd}=\text{even}$, the product of two non-quadratic residues is a quadratic residue, i.e. the Legendre symbol is multiplicative.
