This statement is from the book of Winitzki Linear Algebra via Exterior Products. (Section 3.4, page 123) Let $V$ be finite dimensional vector space, $\dim(V)=N$. The determinant of the matrix $v_{ij}$ is a linear function of each of the vectors $\{\mathbf{v}_i \}$. Consider $\det(v_{ij})$ as a linear function of the first vector, $\mathbf{v}_1$; this function is a covector that we may temporarily denote by $\mathbf{f}_1^{*}$. Show that $\mathbf{f}_1^{*}$ can be represented in the dual basis $\{\mathbf{e}_j^{*} \}$ as $ \mathbf{f}_1^{*}=\sum_{i=1}^N(-1)^{i-1} B_{1i}\mathbf{e}_i^{*}, $ where the coefficients $B_{1i}$ are minors of the matrix $v_{ij}$, that is, determinants of the matrix $v_{ij}$ from which row 1 and column $i$ have been deleted.
Solution. Consider one of the coefficients, for example $B_{11}\equiv\mathbf{f}_1^{*}(\mathbf{e}_1)$. This coefficient can be determined from the tensor equality $ \mathbf{e}_1\wedge\mathbf{v}_2\wedge\ldots\wedge\mathbf{v}_N=B_{11}\mathbf{e}_1\wedge\ldots\wedge\mathbf{e}_N.\qquad(1) $
Edit: the next sentence is
"We could reduce $B_{11}$ to a determinant of an $(n-1)\times(N-1)$ matrix if we could cancel $\mathbf{e}_1$on both sides of equality."
I don't understand the reason of this equality (1). Accepting this equality to be valid, the remainder part of the proof is correct. Please, give some details.