I am doing a project for a graph theory course and would like to prove the Matrix Tree Theorem. This proof uses the Cauchy-Binet formula which I need to prove first. I have found many different proofs of the formula but I am confused about one step. My basic understanding of linear algebra is holding me back.
I am confused about how
$\sum\limits_{1\leq k_1 < ,...,<k_m \leq n} A\begin{pmatrix} 1 & 2 & ... & m\\ k_1 & k_2 & ... & k_m \end{pmatrix} \sum\limits_{\sigma \in S_m} \text{sgn}(\sigma)b_{{k}_{\sigma(1)}1}b_{{k}_{\sigma(2)}2}...b_{{k}_{\sigma(m)}m}$
is equal to
$\sum\limits_{1\leq k_1 < ,...,<k_m \leq n} A\begin{pmatrix} 1 & 2 & ... & m\\ k_1 & k_2 & ... & k_m \end{pmatrix} B\begin{pmatrix} 1 & 2 & ... & m\\ k_1 & k_2 & ... & k_m \end{pmatrix} $
This is the last line step from this proof https://planetmath.org/cauchybinetformula
I understand this is similar to Cauchy-Binet Formula (Matrix Proof)
Thank you in advance.
The answer is that the determinant is expressed using the Leibniz formula for determinants.
https://en.wikipedia.org/wiki/Leibniz_formula_for_determinants