seems to me you can just say
Because AB is given to be invertible
$$(AB)^{-1}=B^{-1}A^{-1}$$
and because A is given to be invertible
$$(AB)^{-1}A=B^{-1}A^{-1}A$$
$$B^{-1}=(AB)^{-1}A$$
So there is a preposition that says;
If AB is invertible, then A is invertible and B is invertible too.
I uploaded the proof that I made, however I think it's not enough.
The problem with my proof, in my opinion, is that I showed that B and A is invertible from one side (either from left or right), so I think that in order to show that they are invertible I have to show that they have inverses from both side and they are the same, am I mistaken?
I would appreciate if you can verify, and tell me if my proof is correct, however if it's not don't tell me what's the right proof, I want to work on it. Thank you!
You can simplify it a great deal:
If AB is invertible, there is a matrix C such that C(AB) = (AB)C = I.
By associativity, we have: (CA)B = I, and A(BC) = I.
You are correct in that you have only shown B is left-invertible, and A is right-invertible.
We have 4 cases to consider:
1) A is invertible, B is not,
2) B is invertible, A is not,
3) Neither are invertible,
4) Both are invertible.
Let's look at case (1):
We know there is a matrix C such that (AB)C = I. Multiplying by A^{-1}, we get:
A^{-1}[(AB)C] = A^{-1}I = A^{-1}.
(A^{-1}A)(BC) = A^{-1}
BC = A^{-1}
(BC)A = A^{-1}A = I
B(CA) = I, showing B is right-invertible, with the same right-inverse as left-inverse, that is: B is invertible, contradicting our assumption. So case 1 cannot occur.
I think you can easily do case (2).
For case (3) show that B singular implies AB singular, which is impossible. What's left?
There is a problem with your proof for case (3), can you see what it is? (Hint: suppose A was the 0-matrix...does it follow from D ≠ I that AD ≠ A?).
You could modify your original proof by using column reduction instead of row reduction, this would give you the inverses on "the other side".