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.    

$\;AB\;$ invertible $\;\implies \exists\;C\;$ s.t.$\;C(AB)=I\;$ , but using associativity of matrix multiplication:
$$I=C(AB)=(CA)B\implies B\;\;\text{is invertible and}\;\;CA=B^{-1}$$
