N-abelian implies n(n-1)-central

Statement
Suppose $$G$$ is a group and $$n$$ is an integer other than 0 or 1. Suppose that $$G$$ is a n-abelian group, i.e., the map $$x \mapsto x^n$$ is an endomorphism of $$G$$ (hence, it is a  universal power endomorphism). Then, $$G$$ is n(n-1)-central: the inner automorphism group of $$G$$ has exponent dividing $$n(n-1)$$.

Converse

 * 2-central implies 4-abelian
 * 3-central implies 9-abelian
 * 4-central implies 16-abelian
 * 6-central implies 36-abelian

Facts about n-abelian groups
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.

Facts used

 * 1) uses::n-abelian implies every nth power and (n-1)th power commute

Proof
Given: A group $$G$$ such that the map $$x \mapsto x^n$$ is an endomorphism of $$G$$. Elements $$g,h \in G$$ (possibly equal).

To prove: $$g^{n(n-1)}$$ commutes with $$h$$. This is sufficient because $$h$$ being arbitrary, it shows that $$g^{n(n-1)}$$ is central, and $$g$$ being arbitrary, it shows that $$G/Z(G)$$ has exponent dividing $$n(n-1)$$.

Proof: