$\operatorname{rank}AB\leq \operatorname{rank}A, \operatorname{rank}B$ 
Prove that if $A,B$ are any such matrices such that $AB$ exists, then $\operatorname{rank}AB \leq \operatorname{rank}A,\operatorname{rank}B$.

I came across this exercise while doing problems in my textbook, but am not sure where to start for the proof of this. I think columnspace might be involved in the proof, although I am not sure.

I think it is best to prove two things that are stronger statements:
$x\in \ker B \Rightarrow x\in \ker AB$,
$y\in \text{im}\;AB \Rightarrow y\in  \text{im}\;A$.
Together with the rank-nullity theorem, I believe this provides a solution not in the link @Amzoti provided above. Edit- nevermind, it's in there, but not all in one place.
