Order of Group Element equals Order of Coprime Power/Proof 2

Proof
Let $\order g = n$.

Necessary Condition
Let $m \perp n$.

Then by Bézout's Identity:
 * $\exists x, y \in \Z: x m + y n = 1$

Let $h = g^m$.

Then:
 * $h^x = g^{m x} = g^{1 - y n} = g g^{- y n} = g e = g$

and so $g$ is also a power of $h$.

Hence from Order of Group Element not less than Order of Power:
 * $\order g \le \order h \le \order g$

and it follows that:
 * $\order g = \order h$

Sufficient Condition
We prove the contrapositive.

Suppose $m \not \perp n$.

Then:
 * $\exists d \in \Z_{>1}: \exists a, b \in \Z: m = a d \land n = b d$

Let $h = g^m$.

Then:
 * $h^b = \paren {g^m}^{n/d} = g^{a n} = \paren {g^n}^a = e^a = e$

Hence:
 * $\order h \le b < n$

and it follows that:
 * $\order g \ne \order h$