Someone pointed out another solution on how to prove Showing the determinant for a specific type of matrix is $\det(A) = (-1)^n 2^{n-1} \sum_{i=1}^{n} a_1 a_2 \ldots a_{i-1} a_{i+1} \ldots a_n$ and it involves a type of cofactor expansion theorem of a specific form.
Assuming some additional conditions on the ring and using the following formula
$\det(A) = \det (\begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ . & . & . & . \\ a_{n1} & a_{n2} & \ldots & a_{nn} \\ \end{pmatrix} ) z - \sum_{i=1}^{n} \sum_{j=1}^{n} (-1)^{i + j} M_{ij}x_i y_i$
where
$ A = \begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} & x_1 \\ a_{21} & a_{22} & \ldots & a_{2n} & x_2 \\ . & . & . & . & . \\ a_{n1} & a_{n2} & \ldots & a_{nn} & x_n \\ y_1 & y_2 & \ldots & y_n & z \end{pmatrix} $
and $M_{ij}$ is the minor of order $n-1$.
Then if we assume all $a_{ij}$ are elements of a field and $\det (\begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ . & . & . & . \\ a_{n1} & a_{n2} & \ldots & a_{nn} \\ \end{pmatrix} ) = 0 $ then
How do you show that $\det(A)$ is the product of a linear form in $x_1, x_2, \ldots, x_n$ and a linear form in $y_1, y_2, \ldots, y_n$?