N-abelian implies every nth power and (n-1)th power commute

Statement
Suppose $$G$$ is a group and $$n$$ is an integer such that $$G$$ is n-abelian: the $$n^{th}$$  power map on $$G$$ is an endomorphism of $$G$$. Then, for every $$g,h \in G$$ we have $$h^ng^{n-1} = g^{n-1}h^n$$.

Applications

 * nth power map is automorphism implies (n-1)th power map is endomorphism taking values in the center
 * nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))
 * n-abelian implies n(n-1)-central

Other related facts
We say that a group is a n-abelian group if the $$n^{th}$$ power map is an endomorphism. Here are some related facts about $$n$$-abelian groups.

Analogues in other algebraic structures

 * Multiplication by n map is a derivation iff derived subring has exponent dividing n
 * Multiplication by n map is an endomorphism iff derived subring has exponent dividing n(n-1)

Facts used

 * 1) uses::Group acts as automorphisms by conjugation: For any $$g \in G$$, the map $$c_g:x \mapsto gxg^{-1}$$ is an automorphism of $$G$$.

From the power map being an endomorphism
Given: A group $$G$$ and a natural number $$n$$ such that $$g \mapsto g^n$$ is an endomorphism of $$G$$.

To prove: For every $$g,h \in G$$, $$h^ng^{n-1} = g^{n-1}h^n$$.

Proof: We denote by $$c_g$$ the map $$x \mapsto gxg^{-1}$$.