It is posslible to do this proof in a direct way using a slightly different and in my opinion a lot more sensible approach to the creation/annihilation operators. I will write it up here for the benefit of any interested.
The inner product of $N$ particle states can be written as: $$ \langle\chi_1,...,\chi_N|\Psi_1,...,\Psi_N\rangle= \det {\begin{pmatrix}\langle\chi_1|\Psi_1\rangle &...& \langle\chi_1|\Psi_N\rangle \.\.\\langle\chi_N|\Psi_1\rangle& ...& \langle\chi_N|\Psi_N\rangle\end{pmatrix}}$$ Using th Slater determinant for fermion- and the permanent for Boson-states.
With this we get a new representation of the annihilation operator $C_a$. 66
We are now looking at $N-1$ particle states:
$$ \langle\chi_1,...,\chi_{N-1}|C_a |\Psi_1,...,\Psi_N\rangle = \langle\Psi_1,...,\Psi_N|C_a^\dagger |\chi_1,...,\chi_{N-1}\rangle^ \ \langle\Psi_1,...,\Psi_N| a,\chi_1,...,\chi_{N-1}\rangle^=\det {\begin{pmatrix}\langle \Psi_1 |a \rangle &\langle\Psi_1|\chi_1\rangle &...& \langle\Psi_1|\chi_{N-1}\rangle \.\.\ \langle \Psi_1|a \rangle & \langle\Psi_N|\chi_1\rangle& ...& \langle\ \Psi_N|\chi_{N-1}\rangle\end{pmatrix}}^*$$ from the definition of the determinant we know we can write this equally as: $$=\sum \limits_{k=1}^N \xi^{k-1}\langle a|\Psi_k \rangle \langle\chi_1,...,\chi_{N-1}|\Psi_1,...,\sim\Psi_k,...,\Psi_N\rangle$$ whre $\xi=-1$ for fermions and $+1$ for bosons and $\sim\Psi_k$ means $\Psi_k$ is not part of that state. This gives a new representation of the annihilatinon operator $$C_a|\Psi_1,...,\Psi_N\rangle=\sum \limits_{k=1}^N \xi^{k-1}\langle a|\Psi_k \rangle |\Psi_1,...,\sim\Psi_k,...,\Psi_N\rangle$$
We now need to look at a one particle Operator $R_{ab}=|a\rangle \langle b|$ definied as in the question above: $$R_{ab}|\Psi_1,...,\Psi_N\rangle=\sum \limits_i^N |\Psi_1\rangle\otimes..\otimes \underbrace{| a \rangle}{\text{i-th ket}} \otimes...\otimes|\Psi_N\rangle \langle b | \Psi_i\rangle $$ with this and the newfound representation of the operators we see that the operator $R{ab}$ can be writtten as $R_{ab}=C_a^\dagger C_b$. How can we see this?
First $$C_a^\dagger C_b |\Psi_1,...,\Psi_N\rangle= \sum \limits_{k=1}^N \xi^{k-1}\langle b|\Psi_k \rangle |a,\Psi_1,...,\sim\Psi_k,...,\Psi_N\rangle$$ second using the symmetry of the states we get: $$|a,\Psi_1,...,\Psi_{k-1},\Psi_{k+1},...,\Psi_N\rangle=\xi^{k-1}|\Psi_1,... 66