In a group $G$, prove the following result Let $G$ be a group in which $a^5=e$ and $aba^{-1}=b^m$ for some positive integer $m$, and some $a,b\in G$. Then prove that $b^{m^5-1}=e$.
Progress 
$$aba^{-1}=b^m\Rightarrow  ab^ma^{-1}=b^{m^2}$$
What will be the next?

You proved that
$$
b^{m^k} = a^k b a^{-k}
$$
(just expand $b^m$ in your formula for $b^{m^2}$, and use induction).
So,
$$b^{m^5-1}=b^{-1}b^{m^5}=b^{-1}a^5 b a^{-5} = b^{-1}ebe = e.$$
