Answer
See the proof below.
Work Step by Step
Since $A$ is invertible, then $AA^{-1}=I$ and then $(AA^{-1})^{T}=I^{T}=I$
$(A^{-1})^{T}A^{T}=I$
Since $A$ is invertible, then $|A|\ne0$ and we know that $|A^{T}|=|A|$
Then $|A^{T}|\ne0$ and thus $A^{T}$ is nonsingular and invertible
So, the equation $(A^{-1})^{T}A^{T}=I$ implies that $(A^{-1})^{T}A^{T} (A^{T})^{-1}=I(A^{T})^{-1}=(A^{T})^{-1}$
Therefore $(A^{-1})^{T} I=(A^{-1})^{T}=(A^{T})^{-1}$
Thus, $(A^{-1})^{T}=(A^{T})^{-1}$
