quantum field theory - Spin-statistics theorem proof details

Recently I have read one book where there was some incomprehensible proof of the Pauli's spin-statistics theorem. I want to ask about a few details of the proof.

First, the author derives commutation (anticommutation) relations like $[\hat {\psi} (x), \hat {\psi} (x')]_{\pm}$ for arbitrary moments of time for scalar, E.M. and Dirac theory cases. He notices that all of them depend on the function $$D_{0} = \int e^{i(\mathbf p \cdot \mathbf (\mathbf x - \mathbf x'))}\frac{\sin(\epsilon_{\mathbf p}(t - t'))}{\epsilon_{\mathbf p}}\frac{d^{3}\mathbf p}{(2 \pi)^{3}}, \quad \epsilon^{2}_{\mathbf p} = \mathbf p^{2} + m^{2},$$ which (as it can be showed) is Lorentz-invariant. For example, it is not hard to show that for fermionic field $$[\Psi (x), \Psi^{\dagger } (x')]_{+} = \left( i\gamma^{\mu}\partial_{\mu} + m\right)D_{0}(x - x').$$

Second, he assumes that for the case of arbitrary integer spin $2n$ there exists a function $\Psi(x)$, for which $$[\hat {\Psi}_{a} (x), \hat {\Psi}^{\dagger}_{b}(x')]_{\pm} = F^{\ 2n}_{ab}\left(\frac{\partial}{\partial x}\right)D_{0}(x - x'),$$ and for the case $s = 2n + 1$ there exists a function $\Psi(x)$, for which $$[\hat {\Psi}_{a} (x), \hat {\Psi}^{\dagger}_{b}(x')]_{\pm} = F^{\ 2n + 1}_{ab}\left(\frac{\partial}{\partial x}\right)D_{0}(x - x'),$$ where $F_{ab}^{\\ k}$ refers to the $\frac{\partial}{\partial x}$ polynomial of rank $k$ and the author (at this stage of the proof) doesn't clarify the sign of commutator.

How can one argue such a generalization from spin $0, \frac{1}{2}$ and $1$ cases on the arbitrary cases of spin value? It is a very strong assumption, because formally it almost proves Pauli's theorem.