Q:

Understanding the proof of $\left ( A' \right )^{-1}=\left ( A^{-1} \right )^{'}$

Assume that $A$ is an $n \times n$ square matrix and is nonsingular. 
To prove that $\left ( A' \right )^{-1}=\left ( A^{-1} \right )^{'}$, it is mentioned in the solution of the lecture notes that $$A'\left ( A^{-1} \right )^{'}=\left ( A^{-1} \right )^{'}A'=I.$$
This is the part that I don't understand. How does showing the above help us prove that $\left ( A' \right )^{-1}=\left ( A^{-1} \right )^{'}$?

A:

If you have a matrix $N$ and you find that $MN = NM = I$ then this automatically implies that $M$ is the unique inverse to your matrix $N$. We denote this by $N^{-1} = M$.
So if you show that $A' \left(A^{-1}\right)' =\left(A^{-1}\right)'A' = I$ then you've shown that $\left(A^{-1}\right)'$ is the inverse to $A'$. But we denote the inverse to $A'$ as $\left(A'\right)^{-1}$. So this means that $\left(A'\right)^{-1} = \left(A^{-1}\right)'$. 

If you still can't quite see it, try giving $A'$ a new name; like so. Let $N = A'$ and $M = (A^{-1})'$. Then you've showed $NM= MN = I$. So this means $N^{-1} = M$. That is $(A')^{-1} = M = (A^{-1})'$.

