Showing AB=0 does not imply either A,B=0, but that singular Ex. 8.5 - Mathematical Methods for Physics and Engineering (Riley)

By considering the matrices
  $$ A = \left( \begin{matrix}
        1 & 0 \\
        0 & 0 \\
        \end{matrix} \right) 
\text{ ,  } B = \left( \begin{matrix}
        0 & 0 \\
        3 & 4 \\
        \end{matrix} \right) 
$$
  show that $AB = 0$ does not imply that either $A$ or $B$ is the zero matrix, but that it does imply that at least one of them is singular.

So, my reasoning was the following:
It's not difficult to compute $AB = \left( \begin{matrix}
        0 & 0 \\
        0 & 0 \\
        \end{matrix} \right) $, in fact it's really even implied in the question.
So, assume that $A, B$ are each non-singular - i.e. they are invertible.
Thus, $A^{-1}AB=B$, and $ABB^{-1}=A$.
But $AB$ is a zero matrix, so $A=B=0$.
Thus proven that the initial assumption $A, B$ are non-singular is false.

Is my reasoning correct? I ask because the 'hints and answers' said simply "Use the property of the determinant of a matrix product."
While I don't expect there to be only one proof, it's a tad disconcerting for it to so flatly suggest a different method - I'm not nearly confident enough in my ability to disregard it.
 A: There's nothing wrong with your proof. It can be argued to be even "more basic" than the determinant one. 
A: No need for a contradiction. Matrix $C$ is singular iff $Cx=0$ for some nonzero $x$. $AB=0$ implies that for any $x\neq 0$, $ABx=0$. If $Bx=0$, you are done and $B$ is singular. If $Bx=y\neq 0$, then $Ay=ABx=0$ and $A$ is singular.
With determinants: $AB=0$ implies $\det(AB)=\det(A)\det(B)=0$, so either $\det(A)=0$ or $\det(B)=0$ or both.
A: You can do a bit better than this: if $AB=0$ then either both matrices are singular, or one of them is zero; of course a zero matrix is singular*. Because  if $AB=0$, then if $A$ is non-singular, then one has $B=A^{-1}AB=A^{-1}0=0$; similarly for $B$ non-singular gets $A=0$.
*If like me, you happen to care not to ignore the existence of matrices without entries (which happens if one or both of their dimensions is$~0$), then this is wrong: the $0\times0$ matrix is both nonsingular (it is the identity) and zero. But note that "$AB=0$ implies $A$ or $B$ is singular" also fails for $0\times0$ matrices. So the reformulation I gave, which remains valid for empty matrices, is in fact better than the original formulation.
A: Your proof is good. A product $AB$ can be the zero matrix with $A$ being invertible (or non-singular): just take $B=0$.
Your assignment is to prove that from $AB=0$ it follows that one among $A$ and $B$ is singular.
Now, if $A$ is invertible, then $AB=0$ implies $B=A^{-1}AB=A^{-1}0=0$, so $B$ is certainly singular. QED
Determinants are surely not needed for this.
You can prove more: if $AB=0$ and both $A$ and $B$ are non zero, then both are singular. Indeed, take a nonzero column of $B$, call it $v$; then $Av=0$ and so $A$ is singular. Then apply the same to $B^TA^T$, showing that $B^T$ is singular.
