Prove that $\gamma^{\frac{q-1}{2}}=-1$ in $\mathbb{F}_q$ when $q$ is an odd prime power and $\mathbb{F}_q^*=<\gamma >$. I have written a proof for this, however I think there may be a simpler way to go about it and am curious for any suggestions. Also I don't fully see where my proof would fail if $q$ was even (other than the fact that $2$ does not divide $q-1$). Here is my proof:
Let $\gamma^t$ for some $0\leq t \leq q-1$ be the additive inverse of $1$ in $\mathbb{F}_q$, so $1+\gamma^t=0$. Then $\gamma^i +\gamma^{t+i}=0$ for all $0\leq i\leq q-2$ and in general we can write the additive inverse of an element as $-\gamma^i=\gamma^{t+i}$.
Then we have 
$\gamma^j=-(-\gamma^j)=-(\gamma^{j+t})=\gamma^{j+2t}$
so $2t\equiv 0 \mod q-1$ and $t$ must be equal to $\frac{q-1}{2}$.
 A: Can we do it this way as well?
Since $\gamma$ generates $\Bbb F_q^\ast$, which has $q -1$ elements, we have
$\gamma^{q - 1} = 1; \tag 1$
now
$(\gamma^{\frac{q - 1}{2}} - 1)(\gamma^{\frac{q - 1}{2}} + 1) = (\gamma^{\frac{q - 1}{2}})^2 - 1 = \gamma^{q - 1} -1 = 0, \tag 2$
by (1); also,
$\gamma^{\frac{q - 1}{2}} \ne 1; \tag 3$
otherwise, $\langle \gamma \rangle = F_q^\ast \tag 4$
would only have $\frac{q - 1}{2}$ elements, that is, the order of $\Bbb F_q^\ast$ would only be $\frac{q - 1}{2}$, not $q -1$; since $q$ is an odd prime,
$\dfrac{q - 1}{2} \ne q -1, \tag 5$
and so $\vert \Bbb F_q^\ast \vert = \frac{q -1}{2}$ is impossible.  Thus (3) binds and we may cancel the factor of $\gamma^{\frac{q - 1}{2}} - 1$ out of (2), leaving
$\gamma^{\frac{q - 1}{2}} + 1= 0, \tag 6$
or
$\gamma^{\frac{q - 1}{2}} = -1. \tag 7$
A: I want to proof by Wilson's theorem. By Wilson's theorem, for prime number $q$, we have  the following relation:
$$
1\times 2 \times 3\times \cdots \times {p-1}=(p-1)! \equiv -1 \pmod{p} \tag{1}
$$
Because  $\gamma$ is the primitive elements of $\mathbb{F}^*$, then every element of this filed can be shown as a power of $\gamma$. Therefore, the relation $(1)$ can be written as the following form:
$$
\gamma^{i_1}\times\gamma^{i_2} \times \gamma^{i_3}\times \cdots \times \gamma^{i_{p-1}}\equiv -1 \pmod{q} \tag{2}
$$ 
Inverse of $\gamma^{i_j}$, $1\leq j \leq q-1$, is $\gamma^{q-1-i_j}$. It is easy to see that every $\gamma^{i_j}$ has a different inverse($\gamma^{i_j}\neq \gamma^{-i_j}$) except two elements, $\gamma^0$ and $\gamma^{\frac{q-1}{2}}$. So the relation $(2)$ can be  changed as follows:
$$
\gamma^{0}\times\gamma^{\frac{q-1}{2}} \equiv -1 \pmod{q} \tag{2}
$$ 
which complete the proof. 
