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.    

Thanking Ted Shifrin I completely revise my answer.
You want to prove that $B$ is invertible using only semigroup properties of matrices (i.e. multiplication, associativity and existence of the unity  $I$). 
You had proved that $B$ has a left inverse, $CB=I$. Now it should prove that there is a right inverse, $BD=I$. However, it is not true in semigroups, generally speaking. An example: so called bicyclic monoid $S=\langle a,b| ab=1\rangle$ (it is generated by $a,b$ with the defining relation $ab=1$). In $S$ the product $ab$ is invertible (since it equals $1$), $b$ has the left inverse $a$, but has not  a right inverse.
Сonclusion: To prove what you want, it is not enough to use  semigroup properties. One have to use either determinants or vectors (or something else).
