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.    
 A: $\;AB\;$ invertible $\;\implies \exists\;C\;$ s.t.$\;C(AB)=I\;$ , but using associativity of matrix multiplication:
$$I=C(AB)=(CA)B\implies B\;\;\text{is invertible and}\;\;CA=B^{-1}$$
A: 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).
A: 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.
A: $n=\mathrm{rank}(AB)\le\mathrm{rank}(B)$, therefore $\mathrm{rank}(B)=n$.
A: A more basic argument, based on invertible $\iff$ nonsingular, is as follows. If $B$ were singular, there would be $x\ne 0$ with $Bx=0$, hence with $(AB)x=0$, whence $AB$ is likewise singular.
A: Do you have the theorem that $X$ is invertible if and only if $\det X \neq 0$?
If so, then if $AB$ is invertible, $\det AB\ne 0$.  But $\det AB = \det A\cdot \det B \ne 0$, so both $\det A$ and $\det B$ are nonzero, and therefore both $A$ and $B$ are invertible.
A: Take some arbitrary vector from the null space from $B$ and call this $\vec{w}$. Than $B\vec{w} = \vec{0}$ and so is $AB\vec{w} = \vec{0}$. We conclude that $\text{null}(B) \subseteq \text{null}(AB)$. Now since $AB$ is invertible we know that $\text{null}(AB) = \{ \vec{0} \}$, and therefore so is $\text{null}(B)$. Using the fundamental theorem of invertible matrices we conclude that $B$ is invertible as well.
A: Let $A,B\in M_{n\times n}(F)$ and let $AB$ be invertible. 
Suppose that $B$ is not invertible. The linear transformation $L_B:F^n\rightarrow F^n$, $L_B(x)=Bx$ is then not invertible as the matrix representation of $L_B$ is not invertible. However, since it is a linear transformation and the domain and codomain are each of the same finite dimension, it follows that $L_B$ is not injective and not surjective. As $L_B$ is not injective, there exists some $x\in F^n$ where $L_B(x)=Bx=0$ and $x\neq 0$.
However, $Bx=0$ means
$$\left[(AB)^{-1}A\right]Bx=\left[(AB)^{-1}A\right]\cdot 0 \,\Rightarrow\, x=0$$
a contradiction. Therefore $B$ is invertible.
As $AB$ is invertible,
$$A\left[B(AB)^{-1}\right]=I_n.$$
Then
$$(AB)^{-1}AB=I_n\,\Rightarrow\, (AB)^{-1}ABB^{-1}=B^{-1} \,\Rightarrow\, (AB)^{-1}A=B^{-1} \,\Rightarrow\, B(AB)^{-1}A=BB^{-1}$$
which means
$$\left[B(AB)^{-1}\right]A=I_n$$
so $A$ is invertible.
