Transpose of inverse vs inverse of transpose I can't seem to find the answer to this using Google. 

Is the transpose of the inverse of a square matrix the same as the inverse of the transpose of that same matrix?
Is $(A^{-1})^T = (A^T)^{-1}$ you ask.

Well
$$\begin{align}
A^T(A^{-1})^T =  (A^{-1}A)^{T} = I^T = I \\
(A^{-1})^TA^T =  (AA^{-1})^{T} = I^T = I
\end{align}
$$

This proves that the inverse of $A^T$ is $(A^{-1})^T$. So the answer to your question is yes.

Here I have used that 
$$
A^TB^T = (BA)^T.
$$
And we have used that the inverse of a matrix $A$ is exactly (by definition) the matrix $B$ such that $AB = BA = I$.
Let the inverse of $A$ be $B^T$ so that $B$ is the transpose of the inverse of$A$. Then by definition of the inverse

$$AB^{T}=I=B^{T}A.$$

Now taking the transpose

$$BA^T=I=A^TB,$$

which proves that $B$ is also the inverse of the transpose of $A$.