Justify: $R$ is commutative 
Let $R$ be a ring. If $(ab)^n=a^nb^n$ holds for all $n$ and all $a, b\in R$, then justify that $R$ is commutative ring.

No idea how to prove it. But also fail to get one counter example. Please help

By assumption
$$
(a(b+1))^2=a^2(b+1)^2=a^2b^2+2a^2b+a^2
$$
and by distrubutive law
$$
(a(b+1))^2=(ab+a)^2=(ab)^2+aba+a^2b+a^2
$$
From this it follows that
$$
a^2b=aba
$$
and this holds for all $a,b \in R$. Especially for $a+1$ and $b$
$$
(a+1)^2b=(a+1)b(a+1)=(ab+b)(a+1)=aba+ab+ba+b
$$
but compute the left side to get
$$
(a+1)^2b=(a^2+2a+1)b=a^2b+2ab+b=aba+ab+ab+b
$$
now compare
