If we have the matrix of individual wavefunctions for a two electron system \[\begin{bmatrix} \phi_1(r_1) & \phi_2(r_1) \\ \phi_1(r_2) & \phi_2(r_2) \end{bmatrix}\]
We know that the antisymmetric wavefunction of that system is \[\Psi(r_1,r_2)= \begin{vmatrix} \phi_1(r_1) & \phi_2(r_1) \\ \phi_1(r_2) & \phi_2(r_2) \end{vmatrix}\]
But from equation \(4\) we know that \(|A|A^{-1}=M\), where \(M\) is some matrix. However, each element of this matrix is also a determinant.
For growing numbers of electrons these states can be expressed in terms of the old states. \[\Psi(r_1)=\phi_1(r_1) \\ \sqrt{2}\Psi(r_1,r_2)=\phi_1(r_1)\phi_2(r_2)-\phi_2(r_1)\phi_1(r_2) = \Psi(r_1)\phi_2(r_2)-\phi_2(r_1)\Psi(r_2) \\ \sqrt{6}\Psi(r_1,r_2,r_3)=\dots=\Psi(r_1,r_2)\phi_3(r_3)+\Psi(r_2,r_3)\phi_3(r_1)+\Psi(r_3,r_1)\phi_3(r_2)\]
This ends with \[\Psi(r_1\dots r_N)=\frac{1}{\sqrt{N!}}\sum_{i=1}^n \Psi(r_{i+1 mod N?},r_{i+2 mod N?}\dots r_{N mod N?})\phi_N(r_i)\]
But we know \[\Psi(r_1,\dots,r_N)=\prod_i \psi_{n_i}(r_i)\]