Raising multiplied group elements to a power If for example we have $(aba^{-1})^n$, how do you go about expanding this to show that it's the same as $a^nba^{-n}=ab^na^{-1}$?
 A: As we discuss in here, it's not (necessarily) the same as $a^nba^{−n}$. We have following Lemma for this case:

Let $aba^{−1}=b^r$. Then $a^nba^{−n}=b^{r^n}$.

Proof: First, we multiply $a$ and $a^{-1}$ from right and left respectively. Thus, one can see that $a^2ba^{−2}=ab^ra^{−1}$. Since $ab^ra^{−1}=(aba^{−1})^r$, $a^2ba^{−2}=(aba^{−1})^r$. Now, we use main relation and so $a^2ba^{−2}=b^{r^2}$. By repeating previous procedure, one can show that $a^nba^{−n}=b^{r^n}$, as desired.
A: Be careful, it's not (necessarily) the same as $a^nba^{-n}$.
$$(aba^{-1})^n = (aba^{-1})(aba^{-1})\ldots (aba^{-1})$$
What happens to all of the $a^{-1}a$ terms that appear next to each other once you apply associativity? To show this with complete rigour, you'll want to use induction.
A: For the latter, use induction. Clearly the equality $(aba^{-1})^n = ab^na^{-1}$ is true for $n = 1$, and by induction hypothesis,
$$(aba^{-1})^n = (aba^{-1})(aba^{-1})^{n-1} = aba^{-1}ab^{n-1}a^{-1}$$
gives you what you want.
The first one is false: Take $a = 1$ and $b \not= b^n$.
