Proving rank of $AB$ is at most equal to rank of $B$ $A=m\times n$ matrix. $B = n\times p$ matrix. Prove that the rank of of the product $AB$ is at most equal to the rank of $B$.
Current state of my work:
(1) First idea: show that the kernel of $B$, $k(B)$, is contained in $k(AB)$. Let $x$ denote an arbitrary element in $k(B)$. Then $ABx = A(Bx) = $A(0) = 0, where 0 is the zero element. Thus, every element of $k(B)$ is in $k(AB)$. Thus the kernel of $B$ is contained in $k(AB)$.
(2) Second idea: Explain why the nullity of $B$ is at most the nullity of $AB$. This follows from (1). If nullity of $B >$ the nullity of $AB$, then there would be some elements of $k(B)$ not expressed in $k(AB)$. But this can't be so since (1) shows that every element of $k(B) \in k(AB)$.
(3) Last idea: Use the rank-nullity theorem. This is where I get stuck. 
The only thing I can come up with is Dim(domain(AB)) = k(AB) + rank(AB). $k(AB) \geq k(B)$. So $Dim(domain(AB)) \geq k(B) + rank(AB)$. 
So I'm stuck. Also, feedback on (1), (2) is appreciated.

Identifying a matrix with its corresponding linear map, the rank of a map $A:\mathbb{R}^n \to \mathbb{R}^m$ is just $\dim (\operatorname{im} A)$. The required statement follows immediately. The rank-nullity theorem allows you make the same argument in terms of $\dim (\ker A)$ rather than $\dim (\operatorname{im} A)$, which looks like the argument in your post.
