If$(ab)^n=a^nb^n$ & $(|G|, n(n-1))=1$ then $G$ is abelian Let $G$ be a group. If $(ab)^n=a^nb^n$ $\forall a,b \in G$ and $(|G|, n(n-1))=1$ then prove that $G$ is abelian.

What I have proven is that:

If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, then $G$ is abelian.

A proof of this can be found in the answers to this old question.

I have got another answer though it can also be easily viewed from James answer too.
We can assume that $n>2$.  Since $(ab)^n = a^nb^n$, for all $a,b\in G$.
Then we  will get $$b^{n-1}a^n = a^nb^{n-1}.$$ this is true for all $a,b\in G$.  
Now see Since the order of $G$ is prime to $n$, the $n$th power map $t\mapsto t^n$ on $G$ is automorphism so is the $(1-n)$ th power map $t\mapsto t^{1-n}$ on $G$.
so there exists $a\in G$ & $b\in G$ such that $x = a^n$ & $y=b^{n-1}$.  Therefore, $xy = a^nb^{n-1} = b^{n-1}a^n = yx$.  Because $x$ and $y$ were arbitrary, it follows that $G$ is commutative.
