But with multiple creation operators we can occupy multiple states: $$ \phi_i(x) \phi_j(x^\prime) \phi_k(x^{\prime \prime}) \rightarrow \hat{a}^\dagger_i \hat{a}^\dagger_j \hat{a}^\dagger_k |0 \rangle. $$

Now we impose antisymmetry, by having the fermion operators satisfy anticommutation relations: $$ \hat{a}^\dagger_i \hat{a}^\dagger_j + \hat{a}^\dagger_j \hat{a}^\dagger_i = [ \hat{a}^\dagger_i ,\hat{a}^\dagger_j ]_+ = \{ \hat{a}^\dagger_i ,\hat{a}^\dagger_j \} = 0 $$ so that $$ \hat{a}^\dagger_i \hat{a}^\dagger_j = - \hat{a}^\dagger_j \hat{a}^\dagger_i $$