Prove that $x^m$ is a generator of the cyclic group $G$ of order $n$ if and only if $(m,n) = 1$

I want to check my proof of the statement.

First, i’ll state the following theorem:

Theorem 1.
Let $o(x)$ be the order of $x in G$. If $o(x)=n$ and $(m,n) = d$ then $o(x^m) = frac{n}{d}$.

Also, a corollary of a theorem (Dan Saracino’s Abstract Algebra, theorem 4.5) that i won’t state here says that:

Corollary 1. Let $langle x rangle$ denote a generator of a group. If $G = langle x rangle$, then $|G| = o(x)$.

Proof. Let $G = langle x rangle$ be a cyclic group of order n. Let’s assume that $x^m$ is a generator of $G$, then $G = langle x^m rangle$. By theorem 1, we know that $o(x^m) = frac{n}{(m,n)}$. And by corollary 1, the order of $G$ must be $o(x^m) = frac{n}{(m,n)}$. But as we know by hypothesis, the order of $G$ is $n$. Then, $n = frac{n}{(m,n)}$. Finally, for this equality to make sense, $(m,n) = 1$.

Thanks!