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.
 A: The issue in what you did is when you wrote 
By definiton, if matrix $\mathbf A$ is the inverse of matrix $\mathbf B$ then $\mathbf B$ is the inverse of $\mathbf A$.
You are getting to the conclusion without a real proof. Better would be to restate the definition, saying that a matrix is invertible if it exists a matrix $B$ such that
$$AB=BA=I.$$
And from there see that $A$ is an invert of $B$ and that it is the only possibility.
A: Well, this can be seen from an abstract group point of view. Your group is the group of invertible $n\times n$ matrices over a field.
An element $g$ of a group is invertible if there exists an element $h$ in the group such that $gh=1=hg$.
First, the invertible is uniquely determined. To see this, let $f,h$ be inverses of $g$. Then $f = f1 = f(gh) = (fg)h = 1h =h$. For this, the inverse of $g$ is denoted by $g^{-1}$.
For each element $g$ of a group, $g=(g^{-1})^{-1}$. To see this, note that $g$ and $(g^{-1})^{-1}$ are both inverses of $g^{-1}$. Since the inverses are uniquely determined, we have $g=(g^{-1})^{-1}$.
A: 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} $ 
A: 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 $(A^{-1})^{-1} = A$.
A: In order to prove that $A^{-1}$ is invertible, we need to show that their exists a matrix $C$ such that $A^{-1}C=I=CA^{-1}$. Take $C=A$ then $A^{-1}C=A^{-1}A=I=CA^{-1}$. Therefore $A^{-1}$ is invertible as it has a inverse.
A: Set of invertible matrices forms a group named general linear group. And for any group element $g$, inverse of it exists. So $g^{-1}$ and the inverse of $g^{-1}$ (notated as $(g^{-1})^{-1}$) exists. 
$$g \cdot g^{-1} = 1$$
Let's multiply this equation with $(g^{-1})^{-1}$ from right (our group is possibly not commutative):
$$g = (g^{-1})^{-1}$$ 
$\square$
Note that this theorem is true for any group, not only general linear groups. This is the power of modern algebra. It categorizes algebraic structures to categories like groups, rings, fields, modules and studies them extensively. Any result about groups (or any other category) is true for all instances of this group (or any instance of the same category). This saves us from repeating the same proof for different algebraic structures. (Also it gives us intuition on the structure of algebra we deal with. Think of concepts like isomorphism etc.)
While you only proved 1 theorem, i proved hundreds of in 5 lines. 
