Determinant Formula for Wedge Product via Universal Property of Exterior Powers

Question

I'm currently learning about differential forms in my analysis class, and I thought I'd dig a bit more into the linear algebra of exterior powers. I've seen the universal property of $\bigwedge^k(V)$: namely, that any alternating multilinear map $f \colon V^k \to W$ factors uniquely over the wedge map to a linear map $\tilde{f} \colon \bigwedge^k(V) \to W$. In my analysis class, the professor said that if we have a collection of $k$ covectors $\omega_1 \cdots \omega_k \in V^*$, their exterior product $\omega_1 \wedge \cdots \wedge \omega_k$ is an alternating multilinear map on $V^k$. Furthermore, if we have $k$ vectors $v_1 \cdots v_k \in V$ then applying our map $\omega_1 \wedge \cdots \wedge \omega_k$ to our vectors gives the determinant of the matrix $A_{i,j} = \omega_i(v_j)$. I'm not entirely sure where this formula comes from. Can one derive this formula from the universal property of $\bigwedge^K(V^*)$?