Given the following formulation of the determinant with Levi-Civita permutation symbols, show that $\det(AB)=\det A \det B$.
$$\det A = \sum\limits_{ij\cdots l}\epsilon_{ij\cdots l} A_{i1}A_{j2}\cdots A_{ln}\,,\,\,\,\,\,\textrm{where A is an }n\times n \textrm{ matrix}$$
I have been trying to show this for so long, but I can't seem to get past a certain point. Here is my work so far.
$$\begin{align*} \det (AB)&=\sum\limits_{ij\cdots l}\epsilon_{ij\cdots l}(AB)_{i1}(AB)_{j2}\cdots(AB)_{ln}\\ &=\sum\limits_{ij\cdots l}\epsilon_{ij\cdots l}\left(\sum\limits_{k_1}A_{ik_1}B_{k_11}\right)\left(\sum\limits_{k_2}A_{jk_2}B_{k_22}\right)\cdots \left(\sum\limits_{k_n}A_{lk_n}B_{k_nn}\right)\\ &=\sum\limits_{ij\cdots l}\epsilon_{ij\cdots l}\left(\sum\limits_{k_1,k_2,\cdots k_n} A_{ik_1}A_{jk_2}\cdots A_{lk_n}B_{k_11}B_{k_22}\cdots B_{k_nn}\right)\\ &=??? \end{align*} $$
Any tips on how to proceed? Have I made a mistake anywhere?