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?
Best Answer
$$ \begin{align*}\det(AB) &= \sum_{ij\cdots l} \epsilon_{ij\cdots l} \sum_{k_1,\dots,k_n} A_{1k_1}B_{k_1i} A_{2k_2}B_{k_2j} \cdots A_{nk_n}B_{k_nl} \\ &= \sum_{k_1,\dots,k_n} A_{1k_1} A_{2k_2} \cdots A_{nk_n} \sum_{ij\cdots l} \epsilon_{ij\cdots l} B_{k_1i}B_{k_1j} \cdots B_{k_nl} \\ &= \sum_{k_1,\dots,k_n} A_{1k_1} A_{2k_2} \cdots A_{nk_n} \det(C_{k_1k_2\cdots k_n}) ,\end{align*}$$ where $C_{k_1k_2\cdots k_n}$ is the matrix whose first row is the $k_1$th row of $B$, whose second row is the $k_2$th row of $B$, ..., and whose $n$th row is the $k_n$th row of $B$.
Hence $$\det(C_{k_1k_2\cdots k_n}) = \epsilon_{k_1k_2\cdots k_n} \det(B).$$ Therefore $$ \det(AB) = \sum_{k_1,\dots,k_n} \epsilon_{k_1k_2\cdots k_n} A_{1k_1} A_{2k_2} \cdots A_{nk_n} \det(B) = \det(A) \det(B) .$$