Moore-Penrose Pseudoinverse: Because $(AA^+)^T = AA^+$, we have that $A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$

Question

I am studying this answer by user "Etienne dM". They claim that, because $(AA^+)^T = AA^+$, we have that $A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$. However, I do not understand how $A^T(Ax-b) = ((AA^+)A)^Tb - A^Tb = 0$ is possible. In particular, it seems like an extra $AA^+$ term comes out of nowhere. I would greatly appreciate it if people would please take the time to explain what's going on here.

Answer 1

The whole point of the MP inverse is that $AA^+A=A$ (and a few other properties). So $$ A^TAx=A^TAA^+b=A^T(AA^+)^Tb=(AA^+A)^Tb=Ab. $$