Let $\Delta$be a determinant function of $N$ dim vector space $V$. Let $|v\rangle$and $\{|v_k \rangle \}^N_{k = 1} $ in vector space $V$. Proof the following eqatuion： $$\sum_{j=1}^N(-1)^{j-1}\Delta(|v\rangle,|v_1\rangle,\cdots,\widehat{|v_j\rangle},\cdots,|v_N\rangle)|v_j\rangle=\Delta(|v_1\rangle,\cdots,|v_N\rangle)|v\rangle$$
$\widehat{|v_j\rangle}$ means that this vector does not exist here.
$\Delta\equiv\sum_\pi\epsilon_\pi\cdot\pi\mathbf{m}$
where $\mathbf{m}$ is a multilinear map
from Hassani, Sadri - Mathematical Physics A Modern Introduction to Its Foundations page 55 Proposition 2.6.9
my proof:
If $\{|v_k \rangle \} ^ N_{k = 1} $is linear dependent, on both sides of the equation of determinant function vectors are linearly dependent, leads to both side of eqatuion equal to $0$.
If $\{|v_k \rangle \} ^ N_{k = 1} $ is linearly independent, because the dimensions of it is equal to $V$, so that they can span $V $ and be seen as a set of base of $V$, making $|v \rangle $ be linear combination of $\{|v_k \rangle \} ^ N_{k = 1} $ so that $|v\rangle=\sum_{i=1}^N\alpha_i|v_i\rangle$. because determinant function's linearity:
$$\Delta(\sum_{i=1}^N\alpha_i|v_i\rangle,|v_1\rangle,\cdots,\widehat{|v_j\rangle},\cdots,|v_N\rangle)=\sum_{i=1}^N\alpha_i\Delta(|v_i\rangle,|v_1\rangle,\cdots,\widehat{|v_j\rangle},\cdots,|v_N\rangle)\notag$$
$$=\sum_{i=1}^N\sum_{j=1}^N(-1)^{j-1}\alpha_i\Delta(|v_i\rangle,|v_1\rangle,\cdots,\widehat{|v_j\rangle},\cdots,|v_N\rangle)|v_j\rangle$$
$$\Delta(|v\rangle,|v_1\rangle,\cdots,\widehat{|v_j\rangle},\cdots,|v_N\rangle)=0$$unless $i=j$ so $$\sum_{j=1}^N(-1)^{j-1}\alpha_j\Delta(|v_j\rangle,|v_1\rangle,\cdots,\widehat{|v_j\rangle},\cdots,|v_N\rangle)|v_j\rangle$$ $$=\sum_{j=1}^N\alpha_j|v_j\rangle (-1)^{j-1}\Delta(|v_j\rangle,|v_1\rangle,\cdots,\widehat{|v_j\rangle},\cdots,|v_N\rangle)=\Delta(|v_1\rangle,\cdots,|v_N\rangle)|v\rangle$$
$(-1)^{j-1}$is due to the complete antisymmetry of the Levi-Chivita symbol in the definition of the determinant function, such that every time the sequential order of any position is swapped in a permutation, there's a minus sign as result.
in other words, $j=1$ no transposition，$(-1)^{0}=1$，when $j=2$transposition$|v_j\rangle$and$|v_{j-1}\rangle$ to get $-\Delta(|v_1\rangle,|v_j\rangle,\cdots,|v_N\rangle)$，when $j=3$transposition $|v_j\rangle$and$|v_{j-1}\rangle$to get$\Delta(|v_1\rangle,|v_2\rangle,|v_j\rangle,\cdots,|v_N\rangle)$...
I am a junior undergraduate physics student, and have not been exposed to rigorous mathematical training. I don't know how to write a good formal proof, and I'm not sure if my proof is reasonable, correct, or readable, so I would like to hear some suggestions, thanks.