Here's what I came up with:

. And we can write k as c·m + d, with d < m.

So we want to show that .

The left hand side obviously simplifies to d.

On the right hand side we get: . The c·m·b vanishes, because of the modulo. And since d < m, d·b mod (m·b) = d·b. So it simplifies to too, which completes the proof.

Hope that helps :)