The Properties of Matrices 
If $A$ and $B$ are a square matrix and $AB=I$ where $I$ is the identity matrix, show that $AB=BA$.
  I solve it, but does my solution correct ?

My Solution:
$$AB=I$$
 multiplying by $B$
$$BAB=BI$$
multiplying by $A$
$$BABA=BIA$$
by using the properties of matrices
$$(BA)(BA)=BA$$
from the last relation $BA=I$ so $$AB=BA$$

In the last line of your solution I think you are assuming $AB=BA$, which is what you want to prove. Shouldn't your last line be $BA=BA$?
Here is an alternative solution to the others given. Since $AB$ is invertible, $A$ is invertible. To see this, denote left multiplication by $L$. We have that $L_{AB}=L_A\circ L_B$. Since $L_{AB}=I$ is invertible, it must be that $L_A$ is injective. But $L_A$ is a map between vector spaces of equal dimension. Hence $L_A$ is invertible so that by definition $A$ is invertible.
It follows $A^{-1}=A^{-1}I=A^{-1}AB=B$. That is $A^{-1}=B$.
