Question:
Is the following true?
If $ \ A$ and $ \ B$ are invertible then $ \ (AB)^T$ is also invertible.
My attempt:
Since $ \ A$ and $ \ B$ are invertible we know $ A^T$ and $ \ B^T$ are also invertible.
Notice that $ \ (AB)^T = B^TA^T$.
Then, $ \ (B^TA^T)(B^TA^T)^{-1} = (B^TA^T)(A^T)^{-1}(B^T)^{-1} = B^T. I.(B^T)^{-1} = B^T.(B^T)^{-1} = I$.
Hence, $(B^TA^T)^{-1}$ is the inverse and so $ \ (AB)^T$ is invertible.
If you already knew that $B^T A^T$ was invertible, then you could stop there; since $(AB)^T = B^T A^T$, the fact that $B^T A^T$ is invertible means $(AB)^T$ is invertible. There's no point in doing any of the calculation in the next line.
If you did not already know that $B^T A^T$ is invertible, then the calculation in the next line is nonsense, because you make use of the term $(B^T A^T)^{-1}$, a thing you do not know actually makes sense. By the previous line, it is the very thing you are trying to prove!
Henceforth, I will assume you are arguing from a stance where it is not yet known that $B^T A^T$ is invertible. A fragment of the calculation you do can be viewed as scratch work — specifically, that if $(B^T A^T)^{-1}$ is something that makes sense, then $(A^T)^{-1} (B^T)^{-1}$ is the inverse of $B^T A^T$, so that you now have a guess as to what the inverse of $(AB)^T$ should be.
I will assume you've already established that $A^T$ and $B^T$ are invertible, so that $(A^T)^{-1}$ and $(B^T)^{-1}$ make sense.
The thing you should do from there is plug $(A^T)^{-1} (B^T)^{-1}$ into the formulas defining the inverse to see if they hold: your goal is to check:
(in the special case that the linear transformations are just square matrices, it suffices to prove just one of these equations)
And the calculation to check these identities can be extracted from the work you've done.