One of definitions of the determinant is:
$\det ({\mathbf C}) =\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \prod_{k=1}^n C_{k \lambda ({k})}})$
I want to prove from this that
$\det \left({\mathbf {AB}}\right) = \det({\mathbf A})\det({\mathbf B})$
What I have so far:
$(AB)_{k\lambda ({k})} = \sum_{j=1}^n A_{kj}B_{j\lambda(k)}$
so we have for the determinant of $\mathbf {AB}$
$\det ({\mathbf {AB}}) =\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \prod_{k=1}^n \sum_{j=1}^n A_{kj}B_{j\lambda(k)}})$
Now I'm not sure how to denote this, but the product of the sum I think is the sum over all combinations of n terms, each ranging from 1 to n, so I'll denote this set of all combinations $C_n(n)$ for n terms each ranging from 1 to n, analogous to the permutation set, but all combinations instead of permutations.
then I get
$\det ({\mathbf {AB}}) =\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \sum_{\gamma \in C_n(n)} \prod_{k=1}^n A_{k\gamma(k)}B_{\gamma(k)\lambda(k)}} )$
then I can at least seperate the product:
$\det ({\mathbf {AB}}) =\sum_{\lambda \in S_n} ({\operatorname {sgn} ({\lambda}) \sum_{\gamma \in C_n(n)} \prod_{k=1}^n A_{k\gamma(k)} \prod_{r=1}^n B_{\gamma(r)\lambda(r)}} )$
I changed the k to an r in one product because it's a dummy variable so I think it doesn't matter, I don't really know if this thing is helpful but this is my attempt at a solution so far.
Thanks to anyone who helps!