$ST$ and $T$ are invertible. Prove that $S$ is invertible? $(ST)^{-1}$ and $T^{-1}$ exist. So how would I actually show that $S$ is invertible? It seems obvious.
Also just curious, it is true that of $A$ and $B$ are invertible 
then $AB$ is invertible. Is it true that if $AB$ is invertible then $A$ and $B$ are invertible?
 A: Hint
What about
$$ST\circ T^{-1}?$$
A: Hint
From Cauchy's theorem we now that
$$
\det (AB) = \det A \cdot \det B
$$
Determinant of invertible matrices is not equal to $0$ or in other words, the matrix is not invertible if and only if the determinant is equal to $0$.  That might help you to answer one of your questions.
A: If we have proved that $AB$ is invertible if and only if $A$ and $B$ are invertible (supposing $A$ and $B$ are square matrices, say $n\times n$), the invertibility of $S$ when $ST$ and $T$ are invertible follows, because $S=(ST)T^{-1}$.
If $A$ and $B$ are invertible, then $(B^{-1}A^{-1})(AB)=I_n$ (and similarly on the other size), so $AB$ is invertible and $(AB)^{-1}=B^{-1}A^{-1}$.
The converse is not as easy. But we can remember that $A$ is invertible if and only if its rank is $n$. Moreover it's easy to show that
$$
\operatorname{rk}(AB)\le\min\{\operatorname{rk} A,\operatorname{rk} B\}.
$$
Thus $AB$ invertible implies that both $A$ and $B$ have rank $n$, so they are invertible.
For non square matrices it can happen that the product is invertible, but it's known that a non square matrix can't have a two-sided inverse. Example
$$
\begin{bmatrix}1 & 1\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=
\begin{bmatrix}2\end{bmatrix}.
$$
A: For the first question, see the hint in Sami's answer. 
Yes, it is always true that if $A,B$ are invertible, then $AB$ is invertible. The inverse of $AB$ is then given by $B^{-1} A^{-1}$. (prove this)
Yes, it is true that if $AB$ is invertible then both $A$ and $B$ are. At least if you assume that both are square of the same size. Then this proof works: $\det(AB)=\det(A) \det(B) \neq 0$.
