Let $A$ be a nonsingular matrix. Show that $A^{-1}$ is also nonsingular and $(A^{-1})^{-1} = A$ To prove that $A^{-1}$ is also singular:
If $(A^{-1})^{-1}$ is singular then the following must be true:
$$
(A^{-1})^{-1}A^{-1} = I
$$
Since we know that $(A^{-1})^{-1} = A$, then
$$
AA^{-1} = I
$$
which is true.
To prove that $(A^{-1})^{-1} = A$ what I did was:
if
$$
(A^{-1})^{-1} = A
$$
then we can say that
$$
((A^{-1})^{-1})^{-1} = A^{-1}
$$
Substituting the first equation into the second, we'd get
$$
(A)^{-1} = A^{-1}
$$
Which would prove that $(A^{-1})^{-1}$ is indeed $A$; however I'm not sure that this method was entirely proper, mathematically speaking, so I was hoping someone could counfirm that what I did was a proper method, and correct me where I went wrong. What I did feels somewhat circular.
 A: Honestly, you're overcomplicating it, and I don't really understand where you were headed (it's not clear to me what you're assuming and what you're trying to prove... so much so that it seems like you're assuming the conclusion).
You get this statement almost for free if you just look closely at the definition of invertibility of a matrix.

Definition.
Let $A\in M_{n\times n}(F)$ be any matrix. We say $A$ is invertible if there is some matrix $B\in M_{n\times n}(F)$ such that $AB=BA=I_n$. In this case, we refer to $B$ as an inverse of $A$.

In words: a matrix is invertible if I can find some matrix such that the two matrices multiplied together gives the identity (regardless of which way they're multiplied).
Wherever you learn the definition of invertibility, it should immediately be followed up by the following basic theorem:

Theorem.
The matrix $B$ in the statement above is unique. More explicitly, suppose $A$ is given, and there are two matrices $B_1,B_2$ such that $AB_1=B_1A=I_n$ and $AB_2=B_2A=I_n$. Then, we have $B_1=B_2$.
Proof.
An immediate computation:
\begin{align}
B_1=B_1 I_n=B_1(AB_2)=(B_1A)B_2=(I_n)B_2=B_2.
\end{align}

Be sure you can justify each equal sign here. Now, because of this uniqueness result, we denote this matrix $B$ via the symbol $A^{-1}$.
Now, to prove your question, note that $AA^{-1}=A^{-1}A=I_n$. That is literally the proof right there that $A^{-1}$ is invertible and the matrix $A$ is its inverse (because by starting with the matrix $A^{-1}$, I have found some matrix (namely $A$) such that their product (in either order) is the identity).
Perhaps it might be clearer if we just use the notation in the definition itself. The equation $AB=BA=I_n$ tells us by definition that $B=A^{-1}$. But if you read it the other way around, it also tells us that $A=B^{-1}$; again, by definition! If you put these two uses of the definition together then we get $(A^{-1})^{-1}=A$.
A: There is a theoretical criterion very useful to determine whether a matrix is invertible or not.
Precisely, we say that $A\in M_{n}(\textbf{F})$ is invertible iff $\det(A) \neq 0$.
Since $A$ is invertible in the present context, it may be claimed that
\begin{align*}
AA^{-1} = I_{n} \Rightarrow \det(AA^{-1}) = \det(I_{n}) = 1 \Rightarrow \det(A)\det(A^{-1}) = 1 \Rightarrow \det(A^{-1}) \neq 0
\end{align*}
hence it results that $A^{-1}$ is invertible as well.
Now it remains to prove that $(A^{-1})^{-1} = A$. To do so, we shall prove the inverse is unique.
Indeed, suppose there is another $B\in M_{n}(\textbf{F})$ which is the inverse of $A$. It would imply that:
\begin{align*}
A^{-1} = A^{-1}I_{n} = A^{-1}(AB) = (A^{-1}A)B = I_{n}B = B
\end{align*}
Based on such fact, we deduce the proposed result:
\begin{align*}
(A^{-1})^{-1}A^{-1} = I_{n} = AA^{-1} \Rightarrow (A^{-1})^{-1} = A
\end{align*}
and we are done.
Hopefully this helps!
