Let $G$ be cyclic of finite order.
Is it true that every $w \in G$ whose order $ord(w)$ is divisible by an integer $n$ satisfies $n=ord(w^{ord(w^n)})$?
(I ask because it's clear that $ord(w^{ord(w^n)})$ divides $n$. The converse does not hold for a general abelian group, for instance in $\mathbb{Z}_6 \times \mathbb{Z}_2$ we have $ord((1,1)^{ord(3,3)}=6 \neq 3$. So I added the condition of $G$ being cyclic and it feels like it should now hold that $n \leq ord(w^{ord(w^n)})$.)
ATTEMPTED SOLUTION: (Can someone verify?)
$G$ is cyclic of finite order $N$ so let it be generated by $g$ of order $N$.
Count the number of solutions of the equation $x^{ord(w^n)}=1$, $x \in G$. From basic cyclic group theory, the number of solutions is $hcf(ord(w^n),N)=ord(w^n)$ because $ord(w^n)$ divides $ord(w)$ which in turn divides $N$ by Lagrange's Theorem. But I can easily list $ord(w^n)$ solutions: $$w^n,w^{2n},w^{3n}, \ldots ,w^{ord(w^n)n}.$$ I haven't proven these are pairwise distinct, but once this is shown, then it will imply $n \leq ord(w^{ord(w^n)})$ so we'd be done. Indeed, suppose that there was some $1 \leq m < n$ such that $w^{ord(w^n)m}=1$. Then $w^m$ is among $w^n,w^{2n},w^{3n}, \ldots ,w^{ord(w^n)n}.$
That is, $w^{m-tn}=1$ for some $1 \leq t \leq ord(w^n)$. Then $n$ divides $ord(w)$ (by assumption) which in turn divides $m-tn$. Therefore $n$ divides $m$, a contradiction.