Why $a^{\frac{q(q-1)}{2}}= -1$ over the finite field $GF(q)$ when q is odd and $a$ is primitive of the field?

Question

Why $a^{\frac{q(q-1)}{2}}= -1$ over the finite field $GF(q)$ when q is odd and $a$ is primitive of the field?

It's true for q that is even since the characteristic of the in this case is 2 and therefore 1=-1, and $$(a^{(q-1)})^{q/2}=1^{q/2}=1=-1$$ But for q that is odd, I'm missing something.

Answer 1

Hint: $(a^{\frac{q-1}{2}}-1)(a^{\frac{q-1}{2}}+1)=a^{q-1}-1=0$

Answer 2

Hint: Write $a^{\frac{q(q-1)}{2}}= a^0 a^1 \cdots a^{q-1}$ and use Wilson's theorem for abelian groups.