You need to find an $\;X\;$ such that $\;A A^T X = I\;$, given that $\;A\;$ is invertible.
Let's calculate what this means, by repeatedly left-multiplying by a matrix that cancels out: \begin{align} (*) \;\;\; \phantom{\equiv} & A A^T X = I \\ \Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;A^{-1}\;$; use $\;A^{-1} A = I\;$"} \\ & A^T X = A^{-1} \\ \Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;(A^{-1})^T\;$"} \\ & (A^{-1})^T A^T X = (A^{-1})^T A^{-1} \\ \equiv & \;\;\;\;\;\text{"simplify using $\;P^T Q^T = (Q P)^T\;$ and $\;A A^{-1} = I\;$"} \\ (**) \;\;\; \phantom{\equiv} & X = (A^{-1})^T A^{-1} \\ \Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;A A^T\;$ -- working our way back to $(*)$"} \\ & A A^T X = A A^T (A^{-1})^T A^{-1} \\ \equiv & \;\;\;\;\;\text{"simplify using $\;P^T Q^T = (Q P)^T\;$, $\;A^{-1} A = I\;$, and $\;A A^{-1} = I\;$"} \\ & A A^T X = I \\ \end{align}
This 'ping pong' argument shows that $(*)$ and $(**)$ are equivalent, so $\;(A^{-1})^T A^{-1}\;$ is the inverse of $\;A A^T\;$.
The other half of the question can be solved in the same way.