Suppose $g{^n}=e$. Show the order of $g$ divides $n$ Suppose $g{^n}$=e. Show the order of $g$ divides $n$.
Would I use Eulers Theorem???;
$a{^{\phi p}}$ $\equiv1 \pmod p$
$a{^{p-1}}\equiv1 \pmod p$
$a{^p}\equiv a\pmod p$
So then I would have 
$g{^n}\equiv g\pmod n$
then I think you use the $\gcd$, which states $\gcd(a,b) = 1$
or 
$a=nq+r$  and $b=nq+r$
which is $a\equiv b\pmod n$??

You don't need Euler's theorem. You just need division with remainder.
Suppose the order of $g$ is $d$ and $g^n=e$.
Suppose $n$ is not a multiple of $d$, then $n=kd+r$ with $0<r<d$.
It follows that $e=g^n=g^{kd}g^r=eg^r=g^r\implies g^r=e,$ contradicting the fact that $d$ is the order of $g$.
