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.

We can assume that $n>2$.  Since $(ab)^n = a^nb^n$, for all $a,b\in G$, we can write $(ab)^{n+1}$ in two different ways:
$$(ab)^{n+1} = a(ba)^nb = ab^na^nb,$$
and
$$(ab)^{n+1} = ab(ab)^n = aba^nb^n.$$
Hence,
$$ab^na^nb = aba^nb^n.$$
Cancel $ab$ on the left and $b$ on the right to obtain
$$b^{n-1}a^n = a^nb^{n-1}.$$
Note that this is true for all $a,b\in G$.  (This says that the $n$th power of any element of $G$ commutes with the $(n-1)$st power of any element of $G$.)
Now let $x,y\in G$ be arbitrary; we want to show that $x$ and $y$ commute.  Since the order of $G$ is prime to $n$, the $n$th power map $t\mapsto t^n$ on $G$ is bijective, so there exists $a\in G$ such that $x = a^n$.  Since the order of $G$ is prime to $n-1$, there exists $b\in G$ for which $y=b^{n-1}$.  Therefore, $xy = a^nb^{n-1} = b^{n-1}a^n = yx$.  Because $x$ and $y$ were arbitrary, it follows that $G$ is commutative.
ADDED:
Lemma. Let $G$ be a group of finite order $m$, and let $k$ be a positive integer such that $(k,m)=1$.  Then the $k$th power map $x\mapsto x^k$ on $G$ is bijective.
Proof.
Since $G$ is finite, it suffices to show that the map $x\mapsto x^k$ is surjective.  To this end, let $g\in G$; we show that $g$ is the $k$th power of some element in $G$.  Since the order of $g$ divides the order of the group $G$, it follows that $(\left| g\right|, k) = 1$.  Therefore, $\langle g\rangle = \langle g^k\rangle$.  Hence, there is an integer $r$ for which $g = (g^k)^r = g^{kr} = (g^r)^k$.  This completes the proof.
