Proof for if $A$ is invertible then $AB$ is invertible First, $A$ and $B$ are square matrices
So to prove if $AB$ is invertible, then $A$ is invertible:
I let $C=(AB^{-1})B$
Then $CA=(AB^{-1})AB=I$
And $C=A^{-1}$
, so A is invertible.
But how do I prove it the other way around?
I'm pretty sure this would require that $B$ also be invertible for it to be true.
Is there a quick way to disprove that if $A$ is invertible then $AB$ is invertible?
 A: The proof that if $A$ and $B$ are invertible, then $AB$ is invertible can be done more elegantly if you know these two results: 
$(1)$. $\det{AB} = (\det (A))*(\det(B)).$
$(2)$. A matrix $B$ is invertible if and only if $\det(B) \neq 0$. 
Proof: Suppose that both $A$ and $B$ are invertible. Then $\det(A) \neq 0$ and $\det(B) \neq 0$. Now by $(1)$, $\det(AB) \neq 0$, so by $(2)$, $AB$ is invertible.
A: I don't quite get your question since the claim is false. What if $B$ is the zero matrix?
Implication of $(AB)^{-1} \implies det(A) \neq 0$ is trivially true
$$(AB)^{-1} = C = B^{-1}A^{-1} = C \implies A^{-1} = BC = B(AB)^{-1}$$
But the point is, the converse does not apply (if $A$ is invertible then $AB$ is invertible). 
A: Consider the square matrices as linear maps $\mathbb{R}^n \to \mathbb{R}^n$. These three conditions are equivalent: 


*

*$A$ is invertible, 

*$\operatorname{im} A = \mathbb{R}^n$, 

*$\operatorname{ker} A = 0$. 


Since $\operatorname{im} AB \subseteq \operatorname{im} A$, we have 
$$ \operatorname{im} AB = \mathbb{R}^n \Rightarrow \operatorname{im} A = \mathbb{R}^n. $$
Due to the equivalent conditions mentioned above: 
$$ AB \text{ is invertible } \Rightarrow A \text{ is invertible}. $$
You could use the third condition to show that $B$ is also invertible. 
