Q:

How can I show that $\det(v_1,v_2,\ldots,v_n)=dx_1\, dx_2\cdots dx_n(v_1,v_2,\ldots,v_n)$?

I wanted to use the definition of a wedge product which says $λ_1λ_2\cdots λ_k(v_1,v_2,\ldots,v_k)=\det(λ_i(v_j))$ with $1<i,j<k$
but I'm not sure if that even can work

A:

This is essentially by the definition you wrote. It is assumed that $x_1, \ldots, x_n$ is a basis, and by definition $dx_1, \ldots, dx_n$ are the dual basis. Thus if you expand out $v_j=a_{j1}v_1+\cdots + a_{jn}v_n$ in this basis, then $dx_i(v_j)=a_{ji}$.
Also in this basis $(v_1, \ldots, v_n)=(a_{ij})$. So since $\det(a_{ij})=\det(a_{ij}^T)=\det(a_{ji})=dx_1\cdots dx_n(v_1, \ldots , v_n)$ we are done.

