## Wilson’s Theorem

Prove that $x^{p^n-1}-1 = \prod_{\alpha \in \mathbb{F}_{p^n}^\times} (x-\alpha)$. Conclude that $\prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = (-1)^{p^n}$. Deduce Wilson’s Theorem: if $p$ is an odd prime, then $(p-1)! \equiv -1$ mod $p$.

Recall that $x^{p^n}-x = \prod_{\alpha \in \mathbb{F}_{p^n}^\times}(x-\alpha)$ by definition, and that $0 \in \mathbb{F}_{p^n}$ is merely the root having minimal polynomial $x$. So $x^{p^n-1}-1 = \prod_{\alpha \in \mathbb{F}_{p^n}^\times} (x-\alpha)$. Comparing constant coefficients, we have $\prod_{\alpha \in \mathbb{F}_{p^n}^\times} (-\alpha) = -1$, so that \$latex $(-1)^{p^n-1} \prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = -1$, and hence $\prod_{\alpha \in \mathbb{F}_{p^n}^\times} \alpha = (-1)^{p^n}$.

Restrict now to the field $\mathbb{F}_p \cong \mathbb{Z}/(p)$ with $p$ odd. Then $\prod_{\alpha \in \mathbb{F}_p} \alpha = (p-1)!$, and $(-1)^p = -1$. Thus $(p-1)! \equiv -1$ mod $p$.

(I’d like to point out that this is a really roundabout way to prove Wilson’s Theorem. The easy(ier) way is to note that in $\mathbb{Z}/(p)^\times$, every element but -1 is distinct from its inverse.)