Right inverse = Moore-Penrose inverse? Let $X=A^*(AA^*)^{-1}$ right inverse for $A$

Definition
  Let $A$ be an $m \times n$ matrix. Matrix $X$ called pseudo-inverse or generalized inverse of $A$ if and only if $X$ satisfying following properties: 
  i.   $AXA=A$ 
  ii.  $XAX=X$  
  iii. $(AX)^*=AX$  
  iv.  $(XA)^*=XA$ 
  with $A^*$ is transpose of $A$ and $X$ called Moore-Penrose inverse of $A$ if and only if $X$ satisfying all properties.

is right inverse satisfying all properties? 
Proof: 
Let $X=A^*(AA^*)^{-1}$ then 
(iv) 
$(XA)^*=(A^*(AA^*)^{-1}A)^*$ 
$=A^*((AA^*)^{-1})^*A$ 
$=A^*((AA^*)^*)^{-1}A$ 
$=A^*(AA^*)^{-1}A$ 
$=XA$  
Next proof (iii) 
$(AX)^*=(AA^*(AA^*)^{-1})^*$ 
$=((AA^*)^{-1})^*AA^*$ 
$=((AA^*)^*)^{-1}AA^*$ 
$=(AA^*)^{-1}AA^*$ 
I stuck on there 
(ii) 
$XAX=A^*(AA^*)^{-1}AA^*(AA^*)^{-1}$ 
$A^*(AA^*)^{-1}AA^*(AA^*)^{-1}=A^*((AA^*)^{-1}AA^*)(AA^*)^{-1}$ 
$=A^*(AA^*)^{-1}$ 
$=X$  
(i) 
$AXA=AA^*(AA^*)^{-1}A$ 
$=(AA^*(AA^*)^{-1})A$ 
$=A$  
Please help me to proof (iii)!
 A: The $m\times n$ matrix $A$ has a right inverse if and only if it has rank $m$. In this case $AA^*$ also has rank $m$, so it is invertible and $A^*(AA^*)^{-1}$ is a right inverse of $A$.
In this particular case it is indeed the Moore-Penrose pseudoinverse of $A$.
Properties 1, 2 and 3 hold for every right inverse of $A$. Indeed, if $AR=I$, then $ARA=IA=A$ and $RAR=RI=R$.
Property 3 is obvious: $AR=I$, so $(AR)^*=I=AR$.
Property 4 is specific of $X$: write $B=(AA^*)^{-1}$, so $X=A^*B$:
$$
(XA)^*=(A^*BA)^*=A^*B^*A
$$
Since $AA^*$ is symmetric (or Hermitian if we're dealing with complex numbers and $A^*$ means the conjugate transpose), also its inverse is, so $B^*=B$. Hence
$$
(XA)^*=(A^*BA)^*=A^*B^*A=A^*BA=XA
$$

A dual result holds when $A$ has a left inverse; in this case $A^*A$ is invertible and the Moore-Penrose pseudoinverse of $A$ is $(A^*A)^{-1}A^*$.
This is important because if $A$ is a generic matrix and $A=LR$ is a full rank decomposition, that is, $L$ has a left inverse and $R$ has a right inverse (both with the same rank of $A$), then the pseudoinverse $A^+$ of $A$ is $A^+=R^+L^+$, with the pseudoinverses of $L$ and $R$ computed as before.
