If $\bf A$ is invertible then $\bf A^{-1}$ is invertible and $\bf (A^{-1})^{-1}=\bf A$. Isn't this proposition self-evident? So the proposition I came across in the book

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

The author provides a bit verbose proof, but my question is, isn't the proposition above self-evident? (Note that the book has already established the uniqueness of the inverse)
The definition of invertible matrix from the same book

A square matrix $\bf A$ is said to be invertible or non-singular if
  there is a matrix $\bf B$ of the same size such that $\bf AB = BA = I$. Matrix $\bf B$ is called the (multiplicative) inverse of A and is denoted by $\bf A^{-1}$.

Simply by looking at the definition (and also keeping in mind that the inverse is unique), you can immediately conclude that $\bf B$ is invertible too and $\bf B^{-1} = A$.
So again, what are we supposed to prove here?
 A: It is definitely self evident if you understand the definition.
As for the proof, I would view it as more an exercise in logic and proof writing than an exercise in linear algebra.  Students often don't have a full understanding of what a proof is supposed to be and it's most evident when you ask them to prove an "obvious" statement.  If you ask them to prove something moderately hard they do fine because their concept of a proof is that it explains why a not obvious thing is true by boiling down the not obvious statement into smaller obvious statements, and they're comfortable with that because that's how they've always approached new math, but once the statement is obvious they don't know what to write.  So here is an opportunity for the author to show students how you setup a straightforward proof with a clear logical structure that relies explicitly on definitions and previously established theorems.
In the case of $(A^{-1})^{-1} = A$.  Think of the definition of an inverse as stating that a certain relationship holds for the ordered pair $(A, B)$ and when that relationship holds we write $B = A^{-1}$.  What the theorem is saying, and what needs to be proved, is that if the relationship holds for the ordered pair $(A, B)$ then it also holds for the ordered pair $(B, A)$.  So you're proving an if-then statement, meaning you should state clearly in the proof (don't assume the reader knows what you intend) that you're assuming the definition holds for $(A, B)$.  Then use that fact, along with some basic logical manipulations, to show that the definition holds for $(B, A)$.  Since the definition is essentially symmetric this is, as you noted, pretty obvious, but it's a good example of how proofs are structured logically.
