I would recommend a proof by induction on n combined with contradiction to prove the induction step.

If n= 1 B^1= 0 gives immediately B= 0. Suppose that, for some k, B^k= 0 implies that B= 0, then if B^(k+1)= 0 and B is NOT 0, then we can divide both sides by B to get B^k= 0, which implies B= 0, a contradiction.