Prove that if $\mathbf A$ is an invertible matrix then $\mathbf A^{-1}$ is invertible and $\mathbf (\mathbf A^{-1})^{-1} = \mathbf A$ I am asked to prove following proposition:

If $\mathbf A$ is an invertible matrix then $\mathbf A^{-1}$ is invertible and $\mathbf (\mathbf A^{-1})^{-1} = \mathbf A$

My attempt:
Let $\mathbf A$ be arbitrary non-singular matrix. It follows that it has inverse, call it $\mathbf B$:
$$\mathbf B = \mathbf A^{-1}$$
By definiton, if matrix $\mathbf A$ is the inverse of matrix $\mathbf B$ then $\mathbf B$ is the inverse of $\mathbf A$. In other words:
$$(\mathbf B)^{-1} = \mathbf A$$
Since $$\mathbf B = \mathbf A^{-1}$$
It follows that
$$(\mathbf A^{-1})^{-1} = \mathbf A $$
Is it correct? 

Although the proposition is quite simple, the proof provided by the book is a bit convoluted, hence I suspect that my proof may have some mistakes.

You can use the definition of an inverse matrix to do it. We know that, if A is a matrix of order $ n $ and it's inverse, then exists B such that: 
$$ AB = BA = I_n $$
And we know that $ B = A^{-1} $, just by notation.
Then it follows that $ A $ is the inverse of $ B $. Just like before, $ A = B^{-1} $, so $ 
A = (A^{-1})^{-1} $ 
