Suppose $A$ and $B$ are $n \times n$ matrices such that $||I-AB||$ is less than $1$, then show that $AB$ is invertible.
I started to prove this by contradiction. Assuming it wasn't invertible so $(I-AB)x=0$ and $||x|| = 1$ thus $x=ABx$. I tried to find a contradiction but didn't have any luck. Any suggestions?