Why is the product of all units of a finite field equal to $-1$? 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!
 A: This is a special case of Wilson's Theorem in a Finite Commutative Group.  See, for instance, pp. 253-255 of these notes for the statement and proof.
Added: Here is a more direct answer to your question.  Write
$a^{(q-2)(q-1)/2}$ as $\left( a^{\frac{q-1}{2}} \right)^{q-2}$.  Now try to convince yourself that
(i) $a^{\frac{q-1}{2}} = -1$ and
(ii) $(-1)^{q-2} = -1$.
For (i), there is a general recipe for the order of a power of a generator in a cyclic group.  If you don't know about this, see e.g. Proposition 250 on page 241 of the same notes.  You should be able to do (ii) easily; for extra credit, you might want to check that $a^{q-2} = a^{-1}$ for all $a \in \mathbb{F}_q^{\times}$.
A: 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
$\displaystyle \prod_{\alpha \in F^{*}} \alpha = -1$. 
A: The only elements of a field so that $x^2=1$ are $1$ and $-1$. Therefore, when you multiply all the non-zero elements together, each element but $1$ and $-1$ will be paired with its inverse. Thus, the product is $-1$.
Note for Fields with Characteristic 2:
In a field of characteristic $2$, $-1=1$, so there is only one element so that $x^2=1$. Thus, the product is still $-1=1$.
A: If the inverse of $x$ is not equal to $x$, pair $x$ with its inverse. Note that if $x$ gets paired with $y$, then $y$ gets paired with $x$. The only objects that do not get paired are the ones that are their own inverse. 
The object $x$ is its own inverse iff $x^2=1$, or equivalently $(x-1)(x+1)=0$. If the characteristic of the field is $\ne 2$, there are two solutions, $1$ and $-1$. The product of any two paired elements is $1$, so the product of all paired elements is $1$.  The product of $1$ and $-1$ is $-1$, so the full product is $-1$.
If the characteristic is $2$, then $-1=1$. The product of the paired elements is $1$, so the full product is $1$. But this is equal to $-1$.
Remark: While I prefer the pairing argument, yours will work. Take for example the case characteristic $\ne 2$. If $w=a^{(q-1)/2}$, then $w^2=1$. But $w\ne 1$, so $w=-1$. Since $q-2$ is odd, it follows that $\left(a^{(q-1)/2}\right)^{q-2}=-1$. 
