I'm trying to show that if $\displaystyle AB - I_n$ is invertible then:
$\displaystyle B(AB-I_n)^{-1}A - I_n = (BA-I_n)^{-1}$
I have no idea where to start...?
Yes that's an interesting exercise:
Let $\displaystyle R$ be a ring, $\displaystyle a$ and $\displaystyle b$ two elements in $\displaystyle R$.
Prove that if $\displaystyle 1-ab$ is left invertible (resp. right invertible), then $\displaystyle 1-ba$ is left invertible (resp. right invertible).
Here, $\displaystyle AB-I$ is invertible, and the inverse of $\displaystyle BA-I$ is given, but you have to verify it is correct.
So just multiply $\displaystyle B(AB-I)^{-1}A-I$ by $\displaystyle BA-I$.
Solution:
$\displaystyle (B(AB-I)^{-1}A-I)(BA-I)$
$\displaystyle =B(AB-I)^{-1}ABA-BA-B(AB-I)^{-1}A+I$
$\displaystyle =B(AB-I)^{-1}ABA-B(AB-I)^{-1}A-BA+I$
$\displaystyle =B((AB-I)^{-1}AB-(AB-I)^{-1})A-BA+I$
$\displaystyle =B((AB-I)^{-1}(AB-I))A-BA+I$
$\displaystyle =BA-BA+I$
$\displaystyle =I$
$\displaystyle B(AB-I)^{-1}A-I$ is a left inverse of $\displaystyle BA-I$, therefore it's the inverse of $\displaystyle BA-I$.
Conclusion: $\displaystyle B(AB-I_n)^{-1}A - I_n = (BA-I_n)^{-1}$