Prove that for every idempotent element a belonging to $A$ (ie such that $a^2 = a$) and for every $b$ belonging to $A$ then $ab = ba$.

Question

Let $A$ be a ring without nonzero nilpotent elements. Prove that for every idempotent element a belonging to $A$ (ie such that $a^2 = a$) and for every $b$ belonging to $A$ then $ab = ba$.

I know that proving that $ab=ba$ is equivalent to proving that $ab-ba=0$ but I try to use the hypothesis and arrive at $ab(1-a)=0$ where I cannot conclude anything because $A$ is not a domain of integrity.

Somebody could help me?