I understand this proof of a related fact: Show that $nullity(B)\leq nullity(AB)$ but I'm not sure if it translates to this due to matrix dimensions. If $A$ and $B$ are square, we can say:
Let $x$ be an element of the null space of $A$. Then $(AB)x=(Ax)B=(0)B=0$ so $x$ is an element of the null space of $AB$, implying that $NS(A)\subseteq NS(AB)$, and hence the desired result. But since the matrices aren't necessarily square, can we adapt this proof using the left null space? ie...
Let $x^T$ be an element of the left null space of $A$. Then $x^T(AB)=(x^TA)B=(0)B=0$, so $x^T$ is an element of the left null space of $AB$, and hence nullity$(A)\leq$ nullity$(AB)$.
I'm not sure if this works because I'm not entirely clear on the relation between the left null space and the null space.