TITLE: Why is the product of all units of a finite field equal to $-1$?
QUESTION [15 upvotes]: Suppose $F=\{0,a_1,\dots,a_{q-1}\}$ is a finite field with $q=p^n$ elements. I'm curious, why is the product of all elements of $F^\ast$ equal to $-1$? I know that $F^\ast$ is cyclic, say generated by $a$. Then the product in question can be written as 
$$
a_1\cdots a_{q-1}=1\cdot a\cdot a^2\cdots a^{q-2}=a^{(q-2)(q-1)/2}.
$$
However, I'm having trouble jumping from that product to $-1$. What is the key observation to make here? Thanks!

REPLY [2 votes]: The $q-1$ elements of $F^*$ are roots of the polynomial $x^{q-1}-1$.  Hence, 
$$-1 = \prod_{\alpha \in F^{*}} (-\alpha) 
= (-1)^{q-1}\prod_{\alpha \in F^{*}} \alpha
\Rightarrow \prod_{\alpha \in F^{*}} \alpha = (-1)^q 
= \begin{cases}-1,&q ~ \text{odd},\\
+1,&q ~ \text{even},\end{cases}$$
but, as André reminded you, $-1 = +1$ if the characteristic of the field is $2$,
and so we need not make special cases but just write that
$\displaystyle \prod_{\alpha \in F^{*}} \alpha = -1$.