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> Given that $R$ is a ring with no non-zero nilpotent elements and has $(ab)^2=a^2b^2$ for all $a,b\in R$, show that $R$ is commutative.
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I have previously shown that, if $R$ is unital and has $(ab)^2=a^2b^2$ for all $a,b\in R$, then $R$ is commutative. This allows me to restrict attention to non-unital $R$, but I haven't been able to get anywhere with that "simplification". The proof in the unital case makes explicit use of the unit element, and it does not seem to suggest anything for this new case.

Some of the things I've tried: various element manipulations; looking for some sort of recurrence between powers of a commutator $ab-ba$; studying the set $S=\{(ab-ba)^k\mid k\in\mathbb{N}\}$.

This is from Herstein's Topics in Algebra, second edition, problem #25 in the supplementary problems of chapter 3. Any hints would be appreciated.

**Edit: proof in the unital case**

If $R$ is unital, evaluate $[a(1+b)]^2$ in two different ways, using the stipulation of the problem, to find $a^2b=aba$. Similarly, $[(1+a)b]^2$ gives $ab^2=bab$. Finally, evaluating $[(1+a)(1+b)]^2$ in two different ways, and canceling terms using those two relations, we end up with $ab=ba$.