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.

If $A$ is an invertible matrix, so there is a matrix $A^{-1}$ such that: $$A.A^{-1} = I$$ where $I$ is identity matrix. Thus, $\text{det}(A.A^{-1}) = \text{det}(A).\text{det}(A^{-1}) = \text{det}(I) = 1$ $\Rightarrow$ $A^{-1}$ is an invertible matrix, because $\text{det}(A^{-1}) \neq 0 $.
Now, there is $(A^{-1})^{-1}$  such that $(A^{-1})(A^{-1})^{-1} = I$ (*).
Multiplying on the left both members of (*) $A$, we have : $A.(A^{-1}).(A^{-1})^{-1} = A.I = A$. 
Then $(A^{-1})^{-1} = A$.
