A group $G$ is called of generalized exponent $n$ if there exists elements $a_1,\dots,a_n \in G$ such that $x^{a_1}\cdots x^{a_n}=1$ for all $x\in G$, where $x^a=a^{-1}xa$. See the following question Generalized identities of (soluble) groups

Every nilpotent group $G$ of generalized exponent $m$ has finite exponent dividing $m^c$, where $c$ is the nilpotent class of $G$. If $m=p$ is prime, is it true that the nilpotent group $G$ has exponent dividing $p^2$. I have proved it for $p\in \{2,3,5,7\}$.