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 $g$.
Let $o$ be the order of $g$. Then $n$ can be written as $oq+r$, with $r\in\{0,1,\ldots,o-1\}$. Therefore\begin{align}e&=g^n\\&=g^{oq+r}\\&=(g^o)^q.g^r\\&=g^r.\end{align} Since $g^r=1$, since $o$ is the order of $g$ and since $r<o$, $r$ can only be equal to $0$. And this means that $o\mid n$.
Consider $A = \{ k \in \mathbb Z : g^k =e \}$. Then $A$ is subgroup of $\mathbb Z$ and so $A=m\mathbb Z$, where $m$ is the smallest positive element of $A$. By definition, $m$ is the order of $g$. Thus, if $n\in A$, then $n$ is a multiple of $m$.
This approach may work for you if you see the subgroups of $\mathbb Z$ before talking about abstract groups.
