Prove that if product of matrices is singular, one of the matrices is singular. I'm having trouble with this proof, it would be much easier to work out the other way it seems.

Let $A$ and $B$ be square matrices of equal size. Prove that if $\det(AB) = 0 =C$ then either $A$ or $B$ must be singular.

I claimed that because $AB$, denoted $C$, is zero then by rule of matrix product that one has to be zero. This is obviously not a real proof
 A: Another alternative is to use matrix algebra and a proof by contradiction.
Let $AB = C$. Assume, to the contrary, that $C$ is singular (i.e. $C^{-1}$ does not exist) and neither $A$ nor $B$ is singular.
Thus it should be possible to construct both $A^{-1}$ and $B^{-1}$ and therefore $D = B^{-1}A^{-1}$.
Premultiplying by $D$, we get $DAB = I = DC$. Hence $D = C^{-1}$, and an inverse for $C$ exists, which is a contradiction.
Therefore, at least one of $A$ or $B$ has to be singular.
A: Hint: $\det(AB) = \det(A)\det(B)$. If $0 = \det(AB) = \det(A)\det(B)$, what can you say about $\det(A)$ or $\det(B)$? How does this relate to singularity?
A: Instead of proving $p\Rightarrow q$, it is almost trivial to prove the equivalent statement $\neg q\Rightarrow \neg p$, that is, if $A$ and $B$ are both nonsingular, then $C$ is nonsingular as well.
Assume that both $A$ and $B$ are nonsingular, that is, there is no nonzero $x$ such that $Ax=0$ or $Bx=0$. Take any $x\neq 0$. Then $Bx\neq 0$ since $B$ is nonsingular. Also, since $A$ is nonsingular, $A(Bx)=Cx\neq 0$. Hence $C$ is nonsingular.
A: Check out this matrix product: 
$$\left[\begin{array}{cc}1&1\\1&1\end{array}\right]
\left[\begin{array}{cc}1&1\\-1&-1\end{array}\right]=
\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$$
