Scalar product of two Slater determinants

Question

this is my first question on MSE so I hope to get it right. I'm trying to prove the following fact about Slater determinants: \begin{equation} \langle \psi_1 \wedge \dots \wedge \psi_N, \phi_1 \wedge \dots \wedge \phi_N \rangle = \det(\langle \psi_i, \phi_j \rangle), \end{equation} where my definition of Slater determinants is the usual one: \begin{equation} \psi_1 \wedge \dots \wedge \psi_N := \frac{1}{\sqrt{N!}}\det(\psi_i(x_j)) = \frac{1}{\sqrt{N!}}\sum_{\sigma \in \Sigma_N}\text{sgn}(\sigma)\psi_{\sigma(1)}(x_1)\dots\psi_{\sigma(N)}(x_N) \end{equation} Here is my attempt:\begin{align*} &\langle \psi_1 \wedge \dots \wedge \psi_N, \phi_1 \wedge \dots \wedge \phi_N \rangle = \frac{1}{N!}\sum_{\sigma, \tau \in \Sigma_N}\text{sgn}(\sigma)\text{sgn}(\tau)\langle \psi_{\sigma(1)},\phi_{\tau(1)}\rangle \dots \langle \psi_{\sigma(N)},\phi_{\tau(N)}\rangle =\\ & = \sum_{\tau \in \Sigma_N}\text{sgn}(\tau)\Bigg(\frac{1}{N!}\sum_{\sigma \in \Sigma_N}\text{sgn}(\sigma)\langle \psi_{\sigma(1)},\phi_{\tau(1)}\rangle \dots \langle \psi_{\sigma(N)},\phi_{\tau(N)}\rangle \Bigg) \end{align*} To get the determinant, the sum in parentheses should reduce to the product $\langle \psi_1,\phi_{\tau(1)}\rangle \dots \langle \psi_N,\phi_{\tau(N)}\rangle$, but I don't see why that should be true: any kind of hint is warmly welcome :)