Moore-Penrose Inverse Formula

Let $$A^+$$ denote the Moore-Penrose inverse of a matrix $$A$$.

Problem. Let $$A$$ and $$B$$ be two compatible matrices where $$B$$ has full row rank. Show that $$AB(AB)^+=AA^+.$$

The verification is easy: Both sides are symmetric and $$AA^+AB=AB=AB(AB)^+AB$$. Of course we cannot cancel $$AB$$ on both sides, but I can't figure out how to start and what the condition $$B$$ has full row rank is used for. Hope anyone has good suggestions.

Update. Here are some of my recent findings but they are not sufficient to solve this problem.

Since $$B$$ has full row rank, we have $$B^+=B^T(BB^T)^{-1}$$ and thus $$BB^+=I$$. Thus, we have $$AA^+=AIA^+=ABB^+A^+.$$ If we can show that $$(AB)^+=B^+A^+$$. Then we are done. Nonetheless, this does not generally hold. We can see here that $$B^+A^+$$ is merely a $$(1,2,3)$$-inverse of $$AB$$:

1. $$ABB^+A^+AB=AA^+AB=AB$$.
2. $$B^+A^+ABB^+A^+=B^+A^+AA^+=B^+A^+$$.
3. $$ABB^+A^+=AA^+$$ is symmetric.

The last condition $$B^+A^+AB$$ is symmetric cannot be verified (at least by me). Of course, if $$A$$ has full column rank, then it would be easier because $$A^+A=I$$ provided $$A$$ has full column rank.

Note that $$AA^+$$ is a projection matrix onto the space $$R(A)$$.
Since $$B$$ has full row rank, we indeed have $$R(AB)=R(A)$$. Thus we must have $$(AB)(AB)^+=AA^+$$.