If I want to prove the above statement, I need two directions.
$\Leftarrow$ Let $(a,n)=1.$ Then $ax+ny=1, x,y\in\mathbb{Z}$ Thus, $1-ax=ny \rightarrow \frac{1-ax}{n}=y$. Therefore, $ax\equiv{1}\mod{n}$. Under addition, then $[ax]n=[n], $ but since $[ax]=[a]$ since $x\in\mathbb{Z}$, then $[a]n=[0]$ Does this imply that $\langle{[a]}\rangle=\mathbb{Z}_n?$
$\Rightarrow$ Let $\langle{[a]}\rangle=\mathbb{Z}_n$ Thus, $\forall[x]\in\mathbb{Z}_n, [x]=[a]y, y=1..n.$ Using the argument above but backwards, can I use that to finish the proof?
Am I even on the right track. I feel that I am but feel I'm not wording it correctly...
Hmm, your notation is confusing you. Normally one writes $\bar{a}$ for the class mod $n$ of $a$ in $\mathbb{Z}_n$. The order of $\bar{a}$ is the smallest natural number $k$ such that $k.\bar{a}\equiv 0$ mod $n$. Observe that $k \leqslant n$. Assume $gcd(a,n)=1$. Then from $n | k.a$ it follows that $ n|k$ and hence $k=n$, so $\bar{a}$ is a generator. The other way around, assume $\bar{a}$ has order $n$ and let $p$ be a prime that divides both $a$ and $n$. Then $\bar{a}.\frac{n}{p} \equiv \frac{a}{p}\bar{n} \equiv 0$ mod $n$, which shows that the order of $\bar{a}$ is smaller than $n$, a contradiction.