Unique nth root implies unique mth root for m dividing n

Statement
Suppose $$G$$ is a group, $$g \in G$$, and $$m,n \in \mathbb{N}$$ are such that $$m$$ divides $$n$$. Suppose $$x \in G$$ is the unique element of $$G$$ satisfying $$x^n = g$$. Then, the equation $$y^m = g$$ has a unique solution for $$y \in G$$, namely $$x^{n/m}$$.

Similar facts

 * c-closed implies local powering-invariant: The first two steps, although proved explicitly here, use a variant of theidea.

Applications

 * Equivalence of definitions of local powering-invariant subgroup

Proof
Given: A group $$G$$, an element $$g \in G$$, $$m,n \in \mathbb{N}$$ are such that $$m$$ divides $$n$$. $$x$$ is the unique element of $$G$$ satisfying $$x^n = g$$. An element $$y \in G$$ such that $$y^m = g$$.

To prove: $$y = x^{n/m}$$.

Proof: