@MikeMiller while we're at it, let's prove Wedderburn's little theorem. If $\Bbb F_{q^n}/\Bbb F_q$ is an extension of finite fields, then the norm map is given by $x \mapsto x^{1+q+\dots+q^{n-1}}=x^{\frac{q^n-1}{q-1}}$. So the kernel of the norm on multiplicative groups is the subset of elements which satisfy $x^{\frac{q^n-1}{q-1}}=1$, of which there can be at most ${\frac{q^n-1}{q-1}}$.
But $\Bbb F_{q^n}^\times$ has $q^n-1$ elements and $\Bbb F_q^\times$ has $q-1$ elements, so the norm $\Bbb F_{q^n}^\times \to \Bbb F_q$ is surjective.