$(aba^{-1}b^{-1})^{n(n-1)}=e$ Let $G$ be a group in which for some integer $n \gt 1$, $(ab)^n=a^nb^n$ for all $a, b \in G$. Show that $(aba^{-1}b^{-1})^{n(n-1)}=e$ for all $a, b \in G$
From $(ab)^n=a^nb^n$, we have writing $n$ times $(ab)(ab)....(ab)=a^nb^n$ which gives $(ba)^{n-1}=a^{n-1}b^{n-1}$. 
Now $(ab)^n=(ab)(ab)^{n-1}=a^nb^n$. $\implies(ab)b^{n-1}a^{n-1}=a^nb^n$. Hence we get $b^na^{n-1}=a^{n-1}b^n.$
Now $(aba^{-1}b^{-1})^{n(n-1)}=((aba^{-1}b^{-1})^{n})^{n-1}=((ab)^n(a^{-1}b^{-1})^n)^{n-1}=(a^nb^na^{-n}b^{-n})^{n-1}=(a^{-n}b^{-n})^{n-1}(a^nb^n)^{n-1}=b^{n-n^2}a^{n-n^2}b^{n^2-n}a^{n^2-n}=b^{n-n^2}a^{n-n^2}b^{n^2-n}a^{n^2-n-1}a=b^{n-n^2}a^{n-n^2}a^{n^2-n-1}b^{n^2-n}a=b^{n-n^2}a^{-1}b^{n^2-n}a$
All i need to do is interchange $a$ and $b$ which i am unable to do.

You proved that all $n$th powers commute with all $(n-1)$th powers. So $b^{n-n^2}a^{n-n^2}b^{n^2-n}a^{n^2-n} = b^{n-n^2}b^{n^2-n}a^{n-n^2}a^{n^2-n}=1$.
