Prove: $\newcommand{\adj}{\operatorname{adj}}$If $A$ is invertible, then $\adj(A)$ is invertible and $[\adj(A)]^{-1}=\frac{1}{\det(A)}A=\adj(A^{-1})$
I can show the left side:
\begin{align*} A^{-1}&=\frac{1}{\det(A)}\adj(A)\\ \implies AA^{-1}&=\frac{1}{\det(A)}A \cdot \adj(A)\\ \implies I&=\frac{1}{\det(A)}A\cdot \adj(A), \end{align*} and, \begin{align*} A^{-1}A&=\adj(A)\frac{1}{\det(A)}A\\ \implies I&=\adj(A)\frac{1}{\det(A)}A. \end{align*} So, $$[\adj(A)]^{-1}=\frac{1}{\det(A)}A.$$
But I'm not sure how to show: $$\frac{1}{\det(A)}A=\adj(A^{-1}).$$