Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible. Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible. 

Here is what I have attempted:

We have $AB$ invertible so there exists an $m\times m$ matrix $C$ such that $C(AB)=(AB)C=I_{m}$. Since matrix multiplication is associative we immediately have a right and left inverse respectively: $(CA)B=A(BC)=I_{m}$. I know the fact that I have square matrices needs to play a roll and gaurentees me that the product will be, in general, defined. I have tried to multiply $CA$ and $BC$ on other sides but that didn't get me anywhere. 

I've asked myself what would happen if they weren't square and its obvious, by definition, that they would not be invertible but I am not seeing a way to use the fact that they are to gaurentee ourselves a general inverse.
Okay, so I figured this out.

We have a left and right inverse for our matrices $B$ and $A$ respectively. Since $A,B$ both correspond to linear transformations from a space of dimension n to another of dimension n we know that they are functions which possess a right and left inverse respectively, and thus are onto and one-to-one respectively. Therefore they are both one-to-one and onto since having one property gives the other when we are dealing with vector spaces of finite and equal dimension. Since $A,B$ both correspond to invertible transformations we know they are invertible. Q.E.D.
If $AB$ is invertible, then there exists $D$ s.t. $ABD=I$

Then $A$ is invertible with inverse $BD$.

It remains to show that $B$ is invertible.

Now we have

$$ABD=I$$
$$\to BD=A^{-1}$$
$$\to B=A^{-1}D^{-1}$$

Then $B$ is invertible as it is the product of invertible matrices...wait, is that given?

QED if yes. Idk o/w.