Contraction of the determinant

Question

In $\mathbb{R}^{n+1}$, we consider an orthonormal basis $e_0,\dots,e_n$ and $\alpha_0,\dots,\alpha_n$ its dual basis. Then, the determinant is $\alpha_0 \wedge \cdots \wedge \alpha_n$. I am seeking a contracted formula for $$ \sum_{\sigma \in \Sigma_{n}} |\sigma|\det(e_0,v_{\sigma(1)},\dots,v_{\sigma(n)}) $$ In terms of differential forms, it is given by $$ \sum_{\sigma \in \Sigma_{n}} |\sigma|\alpha_{\sigma(1)} \wedge \cdots \wedge \alpha_{\sigma(n)}(v_1,\dots,v_n) $$ Does anyone know a way to simplify this? $\Sigma_n$ is the set of permutations of $\{1,\dots,n\}$