Prove that the inverse of the inverse of a matrix A is equal to A

Question

I've written a proof of $ (A^{-1})^{-1} = A $, but I'm worried it's a circular argument.
Please let me know if this is solid. I would also be interested if you have any other proofs of this property. I've seen ones on here that say the property follows directly from the definition of the inverse, but I don't see how.

Proof:
By definition, B is the inverse of A if and only if AB = I and BA = I.

Observe,
$ (A^{-1})^{-1} = A $
$ (A^{-1})^{-1}B = AB $
$ (A^{-1})^{-1}B = I $
And
$ (A^{-1})^{-1} = A $
$ B(A^{-1})^{-1} = BA $
$ B(A^{-1})^{-1} = I $


Thus, $ (A^{-1})^{-1} = A $

Answer 1

By definition, $B$ is the inverse of $A$ if and only if $AB = I$ and $BA = I$.

Observe that $A^{-1}A=I$ and $AA^{-1}=I$. So the inverse of $A^{-1}$ is $A$.