TITLE: Proof If $AB-I$ Invertible then $BA-I$ invertible.
QUESTION [6 upvotes]: I have these problems :

Proof If $AB-I$ invertible then $BA-I$ invertible.
Proof If $I-AB$ invertible then $I-BA$ invertible.

I think I solve it correctly, But I'm not so sure, I'll be glad to receive feedback.

If $AB-I$ invertible then :  $$\det|AB-I| \neq 0 \implies \\ \det|A-I||B| \neq 0 \implies \\ \det|B||A-I| \neq 0 
\implies\\ \det|BA-I| \neq 0$$ 

Therefore $BA-I$ invertible.

If $I-AB$ invertible then :

$$\det|I-AB| \neq 0 \implies \\
\det|I-B||A| \neq 0 \implies \\
\det|I-BA| \neq 0$$
Therefore $I-BA$ invertible.

REPLY [19 votes]: There's a slick way to discover the inverse by first solving the problem for (formal) power series.  
$$\begin{eqnarray} \rm (1-ab)^{-1} &=&\rm 1+ ab + a\color{#c00}{ba}b + a\color{#0a0}{baba}b +\,\cdots\\
&=&\rm 1+ a (1\, +\, \color{#c00}{ba}\ \ +\ \ \color{#0a0}{baba}\,\ +\,\cdots)b\\
&=&\rm 1+ a (1\,-\,ba)^{-1}b\end{eqnarray}\qquad\qquad$$
Simple algebra proves that this formula is universally correct (as in Kaladin's answer).  
At first glance, it seems highly remarkable that such a method should work. Halmos posed the challenge of explaining why this works in one of his popular expositions in Math. Intelligencer. Some explanations are known - see here for more (see also here).