Give an alternative proof of Wilson's theorem ($g$ is a primitive root modulo an odd prime $p$) Show that $(p-1)! = g^{\sum _{k=0}^{p-1} k}$, where $g$ is a primitive root modulo an odd prime $p$, the use it to give an alternative proof of Wilson's theorem.
I was thinking that ${\sum _{k=0}^{p-1} k}$ is of the form $\phi(p-1)(p-1)/2$   right?? and then since $g$ is a primitive root $g^{p}=1$
 A: First of all $g$ is primitive root this means that:
$$\Bbb Z_p^*=\{g^0,g^1,g^2,\cdots,g^{p-2}\}=\{1,2,\cdots,p-1\}$$
hence the product of the elements of the two sets are equals:
$$(p-1)!\equiv 1\cdot 2\cdots (p-1)\equiv g^0\cdot g^1\cdots g^{p-1}\equiv g^{0+1+2+\cdots+p-1}\equiv g^{{\sum _{k=0}^{p-1} k}}$$
(note that adding $g^{p-1}=1$ does not change the product) and as you noticed we have:
$${\sum _{k=0}^{p-1} k}=\frac{p(p-1)}{2}$$ and because $g^p\equiv g$ we have $$(p-1)!\equiv g^{{\sum _{k=0}^{p-1} k}}\equiv g^{\frac{p(p-1)}{2}}\equiv (g^p)^{\frac{(p-1)}{2}}\equiv g^{\tfrac{p-1}{2}}$$
but from here we can conclude that $$(p-1)!\equiv -1\mod p$$ because if $g$ is a primitive root then $ g^{\frac{p-1}{2}} \equiv -1 \mod p$

Edit how to prove that $x\equiv g^{\frac{p-1}{2}} \equiv -1 \mod p$ ? we know that $g^{p-1}\equiv x^2\equiv 1\mod p$ hence $(x-1)(x+1)\equiv x^2-1\equiv 0 \mod p$ so $x\equiv \mp 1\mod p$
but $x \not \equiv 1\mod p$ because the order of $g$ would be less then $p-1$ which is not psooible, so $x\equiv -1\mod p$
