Showing $(ab)^n=a^nb^n$ given $(ab)^2=a^2b^2$. This is part of Exercise 2.1.15 of F. M. Goodman's _"Algebra: Abstract and Concrete"_. > Let $G$ be a group. Suppose $(ab)^2=a^2b^2$ for all $a, b\in G$. Show that $(ab)^n=a^nb^n$ for all $a, b\in G$, $n\in\mathbb{N}$. ## My Attempt: I'm using induction on $n$. Let $a, b\in G$. Obviously $(ab)^1=ab=a^1b^1$ so the result holds for $n=1$. Assume $(ab)^r=a^rb^r$ for some $r\in\mathbb{N}$. Consider when $n=r+1$: we have $$\begin{align} (ab)^{r+1}&=(ab)^rab \\\ &=a^rb^rab \\\ &=a^{r+1}a^{-1}b^rab, \end{align}$$ but I don't know where to go from here.

Since $abab=aabb$ you can get $ba=ab$, that is $a,b$ commute. Now you can rearrange any product between them.