Is my proof that $(A^n)^{-1} = (A^{-1})^n$ correct? I am still learning Linear Algebra at it's basic levels, and I encountered a theorem about invertible matrices that stated that:

If $A$ is an invertible matrix, then for $n=0,1,2,3,..$. $A^n$ is invertible and $(A^n)^{-1} = (A^{-1})^n$.

Now, in attempting to write my proof, I proceeded this way (note that it's not complete):
$$A^n(A^{-1})^n=\prod_{i=1}^nA\prod_{i=1}^nA^{-1}=\prod_{i=1}^n(AA^{-1})=\prod_{i=1}^nI=I$$
Is this line of thinking correct? Well, am just returning to math after a long time of little practice, so I could be wrong.
Based on my comment to Dimitri's answer, would my use of this argument improve my proof?
$$\prod_{i=1}^{n-1}A.(AA^{-1}).\prod_{i=1}^{n-1}=\prod_{i=1}^{n-1}A.(I).\prod_{i=1}^{n-1}=...=A.(AIA^{-1}).A^{-1}=AIA^{-1}=AA^{-1}=I$$
After checking the comments, it seems this last argument gives me a correct proof eventually, and I now see that the problem with my original approach was making the argument that:

$$\prod_{i=1}^nA\prod_{i=1}^nA^{-1}=\prod_{i=1}^n(AA^{-1})$$

Which is not necessarily correct, but like @Srivatsan demonstrates, that approach is not at all wrong since :

Notice that $A$ and $A^{−1}$ commute, so this justifies your proof now

Thanks to everyone for guidance, now I see why collaboration is going to make me love math :D
 A: Your proof isn't really right.  What I mean is, for general matrices $A$ and $B$ the statement
$$ A^n B^n = \prod A \prod B = \prod (AB) $$
may be wrong because $A$ and $B$ need not commute.  I'd try doing this by induction instead.
A: As Dimitri points out, your proof is incomplete. You can make it work in two ways:


*

*You can group the middle $AA^{-1}$ (remember that matrix product is associative). Noting that this equals $I$, the product simplifies to $A^{n-1} (A^{-1})^{n-1}$. You can then use induction to argue that $A^n (A^{-1})^{n}$ is $I$ for all $n$.

*This is slightly more general variant of the above trick. Suppose $A$ and $B$ are commuting matrices (i.e., $AB = BA$), then you can indeed use 
$$
A^n B^n = (AB)^n
$$
guilt-free! (Notice that $A$ and $A^{-1}$ commute, so this justifies your proof now so you can justify the proof this way as well. Keep in mind that some justification is necessary, otherwise the proof is wrong or incomplete.) The proof of this fact also uses similar ideas; see if you can figure it out yourself. 
In fact, if you have an arbitrary product of matrices consisting of $m$ $A$'s and $n$ $B$'s (and no other matrices), then you can show that this product equals $A^m B^n$. For example, if $A$ and $B$ commute, then
$$
B^5ABA^2 B^{3}  = A^{1+2} B^{5+1+3} = A^3 B^9.
$$

A method by induction: 
I leave you to verify that the result is true for the base case $n=0$. For the induction step, assume that $$ (A^{n-1})^{-1} = (A^{-1})^{n-1} .$$
We must now prove the claim for $n$. This follows from the chain of equalities: 
$$
\begin{eqnarray*}
(A^n)^{-1} &=& (A \cdot A^{n-1})^{-1} 
\\ &\stackrel{({a})}{=}& (A^{n-1})^{-1} \cdot A^{-1} 
\\ &\stackrel{({b})}{=}& (A^{-1})^{n-1} \cdot A^{-1} 
\\ &=& (A^{-1})^{n}.
\end{eqnarray*}
$$
Be sure to justify each step, particularly the ones marked $(a)$ and $(b)$. 
A: HINT, observe that:  $$A^n(A^{-1})^n = \underbrace{A\ldots AA}_{\textrm{n times A}} A^{-1}A^{-1}\ldots A^{-1} = A\ldots A(AA^{-1})A^{-1}\ldots A^{-1} =  A\ldots AIA^{-1}\ldots A^{-1}$$
Sorry for the dots, but I didn't find a better way to point the idea out!
A: What I know is that if $A$ and $B$ are invertible then $AB$ is so. Then $A^n$ is invertible , I am agree with (mt_ ) by induction you can prove it:
we know that
$n=1$ is correct $A^{-1}A=I$
if $n=k$ is correct ; $(A^{k})^{-1}A^k=I$
then we should show $(A^{k+1})^{-1}(A^{})^{k+1}=I$
we have  $(A^{k+1})^{-1}A^{k+1}=(A^{k}A)^{-1}A^{k+1}=A^{-1}(A^k)^{-1}A^{k}A=A^{-1}(A)=I$
we know that $(AB)^{-1}=B^{-1}A^{-1}$
