Finding the inverse of a block $2\times2$ square matrix $ \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ where $A$ is a square invertible matrix. The hint I got was to rewrite the original matrix as a product of 3 matrices and use the property for inverse of product of matrices $(XYZ)^{-1} = Z^{-1} Y^{-1} X^{-1}$
$\begin{align} \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}  &=   \begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix}  \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}  \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} 
\end{align}\\ $
Then
$\begin{align} 
\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}^{-1}= \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} ^{-1} \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}^{-1}  \begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix}^{-1} 
\end{align}$
The first 2 matrices on the RHS are invertible but the third is not, so I might have gone down the wrong path with the product of 3 matrices I think.
The actual inverse for  $\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ is $\begin{bmatrix} 0 & A^{-T} \\ A^{-1} & -A^{-1}A^{-T} \end{bmatrix}$ which can be found by using the formula at the end of section 3 in Lu and Shiou. I would like to know how to solve the problem without resorting to using a formula. Thanks for any tips.

Ben Grossmann already pointed out that you have a flaw in the product. You can't factor invertible matrix into a product of matrices some of which are not invertible. This is an easy consequence of Binet-Cauchy formula: $\det(AB) = \det A\cdot\det B$.
There is an approach that doesn't require anything smart to find the inverse. All you have to remember is how you invert a $2\times 2$ matrix:
$$
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix}
d & -b\\
-c & a
\end{pmatrix}.
$$
Now, this formula uses that multiplication in a field is commutative, but let us inspect how far can this idea get us in your case:
$$
\begin{pmatrix}
I & A\\
A^T & 0
\end{pmatrix}\begin{pmatrix}
0 & -A\\
-A^T & I
\end{pmatrix} = \begin{pmatrix}
-AA^T & 0\\
0 & -A^TA
\end{pmatrix}.
$$
Multiply everything on the right by \begin{pmatrix}
-(AA^T)^{-1} & 0\\
0 & -(A^TA)^{-1}
\end{pmatrix} to get
$$
\begin{pmatrix}
I & A\\
A^T & 0
\end{pmatrix}\begin{pmatrix}
0 & -A\\
-A^T & I
\end{pmatrix}\begin{pmatrix}
-A^{-T}A^{-1} & 0\\
0 & -A^{-1}A^{-T}
\end{pmatrix} = \begin{pmatrix}
I & 0\\
0 & I
\end{pmatrix},
$$
that is
$$
\begin{pmatrix}
I & A\\
A^T & 0
\end{pmatrix}\begin{pmatrix}
0 & A^{-T}\\
A^{-1} & -A^{-1}A^{-T}
\end{pmatrix} = \begin{pmatrix}
I & 0\\
0 & I
\end{pmatrix}.
$$
