# If $g$ is a primitive root modulo $p$, then $(p-1)! \equiv g^{p(p-1)/2} \pmod{p}$

Does anyone know how to prove the following theorem:

If $g$ is a primitive root modulo $p$ (and $p$ is a prime number), then
$$(p-1)! \equiv g^{p(p-1)/2} \pmod{p}.$$

Hint: $$\sum_{k=1}^{p-1}k=\frac{p(p-1)}{2}$$

Clear for $p=2$. Assume $p\ge 3$.

$g$ being a primitive root mod $p$ means (by definition) $\text{ord}_p (g)=p-1$, i.e. $p-1$ is the smallest $k\ge 1$ such that $g^k\equiv 1\,\pmod{\! p}$.

$$g^{p-1}\equiv 1\,\pmod{\! p}\iff g^{(p-1)/2}\equiv \pm 1\,\pmod{\! p},$$

but since $1\le (p-1)/2<p-1$, we have $g^{(p-1)/2}\equiv -1\pmod{\! p}$.

Then $g^{p(p-1)/2}\equiv (g^{(p-1)/2})^p\equiv (-1)^p\equiv -1\,\pmod{\! p}$, which is equivalent to $(p-1)!$ by Wilson's theorem.

Above I used $a^2\equiv 1\iff a\equiv \pm 1\,\pmod{\! p}$, which is true because by Euclid's lemma:

$$p\mid a^2-1=(a+1)(a-1)\iff (p\mid a+1\ \ \text{ or }\ \ p\mid a-1)$$