Prove that if $AB$ is invertible then $B$ is invertible.

$\;AB\;$ invertible $\;\implies \exists\;C\;$ s.t.$\;C(AB)=I\;$ , but using associativity of matrix multiplication:

$$I=C(AB)=(CA)B\implies B\;\;\text{is invertible and}\;\;CA=B^{-1}$$

A more basic argument, based on invertible $\iff$ nonsingular, is as follows. If $B$ were singular, there would be $x\ne 0$ with $Bx=0$, hence with $(AB)x=0$, whence $AB$ is likewise singular.

Do you have the theorem that $X$ is invertible if and only if $\det X \neq 0$?

If so, then if $AB$ is invertible, $\det AB\ne 0$. But $\det AB = \det A\cdot \det B \ne 0$, so both $\det A$ and $\det B$ are nonzero, and therefore both $A$ and $B$ are invertible.