Suppose $R$ is a ring and $\exists n \in \mathbb{Z}_{> 0}$ such that $(ab)^n=ab, \forall a,b \in R.$ Then $ab = 0 $ iff $ba=0 \ ?$

Question

Suppose $R$ is a ring and there exists $n \in \mathbb{Z}_{> 0}$ such that $(ab)^n=ab, \forall a,b \in R.$ Could anyone advise me on how to prove/disprove that $ab = 0$ iff $ba=0, \forall a,b \in R \ ?$

Hints will suffice, thank you.

Answer 1

Suppose that $ab=0$. Then $ba=(ba)^2=b(ab)a=0$. Conversely, if $ba=0$, then $ab=(ab)^2=a(ba)b=0$.

Answer 2

A thermonuclear way to do his is to notice that the hypothesis implies that for all $a\in R$ there is an $n>1$ such that $a^n=a$ and to remember that this implies, according to a famous theorem of Jacobson, that $R$ is commutative.