TITLE: Proof of Wilson's theorem via finite fields?
QUESTION [7 upvotes]: $(p-1)! = -1 \bmod p$ where $p$ is prime.
I know that there are a lot of ways to prove it with all beauty of different sides of math:
1) Elementary way (to divide in pairs where $x_1 x_2 = \pm 1$)
2) To use Fermat's little theorem and Lagrange theorem with try to prove $x^{p-1} - 1 = (x-1)\dots(x-(p-1)))$
But my teacher wants me to prove it using theory of the finite Galois Fields. He wants me to use material which we study now. 
We studied on our last lectures:

A set of residue classes
The decomposition field of the polynomial $x^{p^m} - x$
Cyclicity of the multiplicative group of the field
Giving the field by the root of an irreducible polynomial
Structure of finite fields
Isomorphism of the Galois fields
Automorphism of the Galois field
Representation of the Galois field by matrices

Maybe this proof is on this site, but I can't find it. And I can't think of proof based on things that I mentioned. 
Sorry if it is some kind off-topic. 
But does anybody know other proof based on mentioned material?
Hint from teacher: consider constant term of $x^{p-1} -  1 $, but it can't be the second proof, because it is based on theory of groups at all,  the previous chapter of our lectures, not the theory of the finite Galois field, current chapter of our lectures.

REPLY [4 votes]: My guess is that you may be expected to use cyclicity of $\Bbb{F}_p^*$. Let $g$ be a generator of this cyclic group of order $p-1$. Then there are two ways of listing the elements of the multiplicative group (in different orders):
$$
\Bbb{F}_p^*=\{1,2,3,\ldots,p-1\}=\{1,g,g^2,\cdots,g^{p-2}\}.
$$
This implies that in $\Bbb{F}_p$ we have
$$
\begin{aligned}
(p-1)!&=1\cdot2\cdot3\cdots(p-1)\\
&=1\cdot g\cdot g^2\cdots g^{p-2}\\
&=g^{0+1+2+3+\cdots+(p-2)}.
\end{aligned}
$$
In the last form the exponent is
$$
E=\sum_{j=0}^{p-2}j=\frac12(p-2)(p-1).
$$
Because $p$ is odd, we see that $E$ is an odd multiple of $(p-1)/2$.
On the other hand, by cyclicity, $g^{(p-1)/2}$ is of order two. As $-1$ is the only element of order two, we thus have
$$
W:=g^{(p-1)/2}=-1,
$$
and
$$
g^E=W^{p-2}=(-1)^{p-2}=-1.
$$