How to show that $B^{-1}\cdot A^{-1}=B\cdot A$? I can't find the solution of this problem:
Given two $n\times n$ square matrices $A,B$ such that $A^2\cdot B^2=I_n$, show that  $B^{-1}\cdot A^{-1}=B\cdot A$.
Thanks in advance.
 A: $A^2 \cdot B^2 = I \Rightarrow A, B$ are invertible. The set of all invertible matrices forms a group. So we also have $B^2 \cdot A^2 = I.$ Hence $B \cdot A = B^{-1}\cdot A^{-1}.$
I'm assuming that the entries of the matrices are from a field.
EDIT: $A^2 \cdot B^2 = I \Rightarrow \text{det}(A^2) \cdot \text{det}(B^2) = \text{det}(A^2 \cdot B^2) = 1 \Rightarrow \text{det}(A^2) \neq 0.$ Now $(\text{det}(A))^2 = \text{det}(A^2) \Rightarrow \text{det}(A) \neq 0.$ Similarly for $B.$
A: Since $A^{2}B^{2}=I_{n}$ note that taking $A^{-1}=ABB$ we have that
$A\cdot A^{-1}=I_n$. For square matrices this is enough to show $A$
is invertible. The same follows for $B$ by taking $B^{-1}=AAB$. Using the facts that $A^{-1}\cdot A=I_{n}$ and $B\cdot B^{-1}=I_{n}$
we get that 
\begin{eqnarray*}
A^{2}B^{2}=I_{n} & \iff & AB^{2}=A^{-1}I_{n},\mbox{ by left multiplication}\\
 & \iff & AB=A^{-1}I_{n}B^{-1},\,\mbox{ by right multiplication}\\
 & \iff & AB=A^{-1}B^{-1}.
\end{eqnarray*}
Taking inverses of both sides and using the property $(AB)^{-1}=B^{-1}A^{-1}$ gives the desired result.  
