Prove (w/o induction) that B^n and 0 are congruent mod B.
Hint from professor: let n = m + 1
Hint isn't really necessary as long as $\displaystyle n \ge 1$, which i think you are assuming
note that $\displaystyle B^n \equiv 0 ~(\bmod ~B) \Longleftrightarrow B \mid (B^n - 0)$
this is obviously the case.
So begin by saying $\displaystyle B^n = B^{m + 1} = B \cdot B^m$
where can we go with this?
You did say without induction this time, didn't you?
B is congruent to 0, mod n, if and only B is a multiple of n: say B= na where a is some integer.
If B is congrunt to 0 mod n, then B= na. $\displaystyle B^m= (na)^B= n^ma^m= n(n^{m-1}a^n)$ is sufficient.
And, of course, 0= 0n.