Prove that if $AB$ is invertible then $B$ is invertible. I know this proof is short but a bit tricky. So I suppose that $AB$ is invertible then $(AB)^{-1}$ exists. We also know $(AB)^{-1}=B^{-1}A^{-1}$. If we let $C=(B^{-1}A^{-1}A)$ then by the invertible matrix theorem we see that since $CA=I$(left inverse) then $B$ is invertible. Would this be correct?
Edit
Suppose $AB$ is invertible. There exists a matrix call it $X$ such that $XAB=I$. Let $C=XA$ Then $CB=I$ and it follows that $B$ is invertible by the invertible matrix theorem.    

This might be easier to do by thinking of $A$ and $B$ as linear operators rather than by matrix arithmetic. Assume $AB$ is invertible and assume that $B$ is not. Thus, $B$ either fails to be injective or fails to be surjective. If $B$ is not injective, then there is $x,y$ with $x \neq y$ such that $Bx=By$, and hence $ABx=ABy$, and $AB$ is not injective, which is a contradiction. Now, assume that $B$ is not surjective. Thus, its image must have dimension strictly less than the dimension of its domain. (Here is where we assume $A,B$ are square.) Thus, the composition $AB$ must have an image of dimension strictly less than the dimension of the domain of $B$, and thus  cannot be surjective, another contradiction. Thus $B$ is injective and surjective and thus invertible.
