$$X=\left(\matrix{A & B\\C & D}\right)$$
where $A,B,C,D$ are all $n \times n$ matrices. Assuming that all stated inverses exist, show that
$$X^{-1}=\begin{pmatrix}(A-BD^{-1}C)^{-1}&(C-DB^{-1}A)^{-1}\\(B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1}\end{pmatrix}$$
It has something to do with Schur complements. But all I can think of is to prove that each element of $X^{-1}$ is the same as the standard block inverse formula but when I do that I keep running back around in circles as this formula contains one of the stated inverses. Any help would be great, thanks ahead of time!
The top left block can be obtained as a Schur complement. The other blocks can be obtained by considering $J(XJ)^{-1}$ (bottom left), $(JX)^{-1}J$ (top right) and $J(JXJ)^{-1}J$ (bottom right), where $J=\pmatrix{0&I\\ I&0}$. For instance, using the result that the top left block of $X^{-1}$ is $(A-BD^{-1}C)^{-1}$, we have $$X^{-1}=J(XJ)^{-1}=J\pmatrix{B&A\\ D&C}^{-1}=J\pmatrix{(B-AC^{-1}D)^{-1}&\ast \\ \ast&\ast}=\pmatrix{\ast&\ast \\ (B-AC^{-1}D)^{-1}&\ast}.$$