How to prove that if A and B are matrices, then the rank of AB <= the rank of B So far I know that when matrices A and B are multiplied, with B on the right, the result, AB, is a linear combination of the columns of A, but I'm not sure what to do with this. 

If you know the rank-nullity theorem, it's easy :
$\forall x \in \text{Ker} B, ABx = A(Bx) = A(0) = 0$
So $x\in \text{Ker} AB$, hence $\text{Ker} B \subset \text{Ker} AB$, that imply that 
$\text{dim Ker} B \leq \text{dim Ker} AB$ , and by the rank nullity theorem, you get that $\text{rank} B \geq \text{rank} AB$
