Let $G$ be a group and $a,b \in G$. If $b^{-1}ab=a^k$, prove that $b^{-r}a^sb^r=a^{sk^r}$.
I see that $a^k$ and $a$ are conjugates in $G$. I could only go as far as the following:
$$b^{-1}ab=a^k \implies b^{-1}a^sb=a^{sk} \implies b^{-r}a^sb^r=b^{-(r-1)}a^{sk}b^{r-1}.$$
Any idea what to do next?
Let $G$ be a group and $a,b \in G$. We know that $b^{-1}ab=a^k$. This implies that $b^{-1}a^sb=a^{sk}$. Then $b^{-1}a^{sk}b=a^{sk^2}$ and $b^{-r}a^sb^r=b^{-(r-1)}a^{sk}b^{(r-1)}$. We assume $b^{-(m-1)}a^{sk}b^{(m-1)}=a^{sk^m}$, where $m \in \mathbb{Z}^{+}$. Then $b^{m}a^{sk}b^m=b^{-1}a^{sk^m}b=a^{sk^{(m+1)}}$. Hence, by principle of mathematical induction, $b^{-(r-1)}a^{sk}b^{(r-1)}=a^{sk^r}$, for any $r \in \mathbb{Z}^{+}$. Hence, $b^{-r}a^sb^r=b^{-(r-1)}a^{sk}b^{(r-1)}=a^{sk^r}.$