Prove (w/o induction) that B^n and 0 are congruent mod B.
Hint from professor: let n = m + 1
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. is sufficient.
And, of course, 0= 0n.