Proving determinant product rule combinatorially 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!
 A: From 
$$\det ({\mathbf {AB}})
      =\sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \sum_{\gamma \in C_n(n)} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left(\prod_{r=1}^n B_{\gamma(r)\lambda(r)} \right)$$
reorder the sums and factor out the first product :
$$\det (\mathbf {AB})
      =\sum_{\gamma \in C_n(n)} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} \right)$$
For $\gamma \in C_n(n)$, suppose that $\gamma$ is not a permutation : there are two indices $i,j$ such that $\gamma(i) = \gamma(j)$.
Let $\tau$ be the transposition swapping $i$ and $j$. In particular, $\gamma \circ \tau = \gamma$.
For any $\lambda \in S_n$, by reordering the product, we get :
$$\operatorname {sgn} (\lambda)\prod_{r=1}^n B_{\gamma(r)\lambda(r)} = \operatorname {sgn} (\lambda)\prod_{r=1}^n B_{\gamma(\tau(r))\lambda(\tau(r))} = - \operatorname {sgn} (\lambda \circ \tau)\prod_{r=1}^n B_{\gamma(r)\lambda \circ \tau(r)} $$
Thus, by duplicating and reorganizing the sum,
$$ \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} = \frac 12 \sum_{\lambda \in S_n} \left( \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{\gamma(r)\lambda(r)} + \operatorname {sgn} (\lambda \circ \tau) \prod_{r=1}^n B_{\gamma(r)\lambda \circ \tau(r)(r)}\right) = 0$$
Therefore we may only keep the $\gamma$ that are permutations, and by reordering the second product, we have :
$$\det (\mathbf {AB})
      =\sum_{\gamma \in S_n} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{r\lambda\circ\gamma^{-1}(r)} \right)$$
By reorganizing the second sum, we get :
$$\det (\mathbf {AB})
      =\sum_{\gamma \in S_n} \left(\prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \operatorname {sgn} (\gamma)\prod_{r=1}^n B_{r\lambda(r)} \right) $$
We can factor out $\operatorname{sgn}(\gamma)$ from the second sum, then factor out the whole second sum from the first sum, to get
$$\det (\mathbf {AB}) = \left( \sum_{\gamma \in S_n} \operatorname {sgn} (\gamma) \prod_{k=1}^n A_{k\gamma(k)} \right) \left( \sum_{\lambda \in S_n} \operatorname {sgn} (\lambda) \prod_{r=1}^n B_{r\lambda(r)} \right) = \det (\mathbf {A})\det (\mathbf {B})$$
