Problem about mp inverse Show that $ABB^+(ABB^+)^+=AB(AB)^+$.
This is an exercise from Matrix Differential Calculus with Applications in Statistics and Econometrics. There are no other assumptions. I know if $|A|\neq 0$ , then $(AB)^+=B^+(ABB^+)^+$, and the above equation establishes. 
 A: What precisely is the question?
You provide that for $\mathbf{A}\ne\mathbf{0}$, then
$$
%
\left( \mathbf{AB} \right)^{+} =
%
\mathbf{B}^{+}
\left( \mathbf{ABB}^{+} \right)^{+}
%
\tag{1}
$$
You note an immediate consequence of $(1)$ is
$$
%
\left( \mathbf{AB} \right)
\left( \mathbf{AB} \right)^{+} =
%
\left( \mathbf{AB} \right)
\mathbf{B}^{+}
\left( \mathbf{ABB}^{+} \right)^{+} =
%
\left( \mathbf{ABB}^{+} \right)
\left( \mathbf{ABB}^{+} \right)^{+}
%
\tag{2}
$$
which seems to be the result you are looking for.
A: For any real matrix $X$, $XX^{+}$ is the orthogonal projection matrix onto the column space of $X$.
It follows consequently that the LHS is the orthogonal projection matrix on the column space of $ABB^{+}$ and the RHS is the orthogonal projection matrix on the column space of $AB$.
So it is enough to show that $ABB^{+}$ and $AB$ have the same column space.
Clearly column space of $ABB^{+}$ is contained in column space of $AB$.
Now $AB = ABB^{+}B$, so the column space of $AB$ is also contained in the column space of $ABB^{+}$. Hence $AB$ and $ABB^{+}$ have the same column space and the result follows.
