Below is proof of the Cauchy-Binet Formula found on Wikipedia.
Why are the last two equations equal?
Let $\textbf{A}$ be a $m\times n$ matrix and $\textbf{B}$ be a $n\times m$ matrix.
$\begin{align*}
\displaystyle \det \left({\mathbf A \mathbf B}\right)&=\sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \epsilon \left({l_1, \ldots, l_m}\right) \left({\sum_{k \mathop = 1}^n a_{1 k} b_{k l_1} }\right) \cdots \left({\sum_{k \mathop = 1}^n a_{m k} b_{k l_m} }\right)\\
&=\sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \epsilon \left({l_1, \ldots, l_m}\right) b_{k_1 l_1} \cdots b_{k_m l_m}\\
&=\sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \det \left({\mathbf B_{k_1 \cdots k_m} }\right)\\
&=\sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} \epsilon \left({k_1, \ldots, k_m}\right)a_{1 k_1} \cdots a_{m k_m} \det \left({\mathbf B_{j_1 \cdots j_m} }\right) \tag{1}\\
&=\sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} \det \left({\mathbf A_{j_1 \cdots j_m} }\right) \det \left({\mathbf B_{j_1 \cdots j_m} }\right) \tag{2}\\
\end{align*}$
This is how far I got starting with equation $(2)$…
$\begin{equation}
\sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} \det \left({\mathbf A_{j_1 \cdots j_m} }\right) \det \left({\mathbf B_{j_1 \cdots j_m} }\right)
=
\sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} (\sum_{1\leq k_1,…,\ k_m\leq n}\epsilon(k_1, k_2, …,k_m)a_{j_1,k_1}a_{j_2,k_2}…a_{j_m,k_m}) \det \left({\mathbf B_{j_1 \cdots j_m} }\right)
\end{equation}$
Best Answer
The last $=$ just factors out $\det A_{j_1\cdots j_m}$, defined in terms of the Levi-Civita symbol.