Let $(A,+,\cdot)$ be a finite ring, and $a,b \in A$ such that $(ab-1)b = 0$. Prove that $b(ab-1) = 0$.
I already have one solution for this problem and it involves using the fact that if $(A,+,\cdot)$ is finite, then $\exists k,m \in \mathbb{N}^*$ such that $b^k = b^m$ (we need to show that $bab = b$, so if $b = b^m$, $bab = bab^m = bab^2b^{m-2} = bbb^{m-2} = b^m = b$, because the hypothesis means that $ab^2 = b$; the solution involves using the hypothesis in proving that $\exists m\in \mathbb{N}, m>1$ such that $b = b^m$).
I am looking for a different solution, because this one seems rather "forced". I have solved other problems like this (for instance: $(M, \cdot)$ a finite monoid, then if $ab=1 \implies ba=1$) by assuming that the conclusion is false, then defining a function on $A$ (in the monoid example, $f:M \to M, f(x) = xa$ ) that is injective or surjective, and because $A$ is a finite set, then it is bijective, and thus I achieve a contradiction.
This kind of solutions seem more natural, at least for me, because in cases like this, when I have a relation between two elements in a ring, I try and "play" with them and achieve other relations involving them that will be useful in proving the problem.
I've tried constructing such a function for this problem, but I didn't succeed. It's harder on this particular problem because I can't prove that functions such as $f(x) = x(ax-1), f(x) = bx, f(x) = bxb$ (or others that are involving the hypothesis, but are using only the second operation of the ring) are injective or surjective, because multiplying by $b$ or $a$ and using the hypothesis, I get that $0=0$, which doesn't prove anything.
Can this problem be solved using this idea, or any other idea, without using the fact that in a finite ring every element $x \in A$ has two powers, $c,d \in \mathbb{N}^*$ such that $x^c = x^d$ ?
Any help (tips, solution) is appreciated.