I have the same question as the person here: Action of Fermionic Creation and Annihilation Operators
The question actually wasn't anwered, because using anticommutation relations between creation $c_\nu^\dagger$ and annihilation $c_\nu$ operators means knowing their explicit form.
Books, that I read, just give the following definitions without any remark: $$c_\nu^\dagger\left | n_1,n_2,\dots,n_\nu,\dots \right \rangle=\left ( -1 \right )^{\sum_{i<\nu}n_i}(1-n_\nu)\left | n_1,n_2,\dots,n_\nu+1,\dots \right \rangle$$ $$c_\nu\left | n_1,n_2,\dots,n_\nu,\dots \right \rangle=\left ( -1 \right )^{\sum_{i<\nu}n_i}n_\nu\left | n_1,n_2,\dots,n_\nu-1,\dots \right \rangle$$
The sense of the multipliers $(1-n_\nu)$ and $n_\nu$ is clear: there can be only $1$ or $0$ fermions in the $\nu$-th one-particle state.
The question is how to obtain the factor $\left ( -1 \right )^{\sum_{i<\nu}n_i}$ without knowing anticommutation relations between $c_\nu$ and $c_\nu^\dagger$?
I found a hint in the book "Feynman Diagram Techniques in Condensed Matter Physics" of Radi A. Jishi (chapter 3.4.1). He uses many-particle state representation by the Slater determinant $\left | \phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle$, where $N$ is current number of fermions and $\phi_{\nu_i}\in\left \{ \phi_1,\phi_2,\dots,\phi_\nu,\dots \right \}$ is a one-particle state of the $i$-th fermion. He claims that: $$c_\nu^\dagger\left | \phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle=\left |\phi_{\nu}\phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle\quad \text{if $\nu_i\neq\nu\quad \forall i$}$$ $$c_\nu\left |\phi_{\nu} \phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle=\left |\phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle$$ (a one-particle state on the new fermion $\phi_{\nu_{N+1}}=\phi_{\nu}$)
It means that $c_\nu^\dagger$ adds the first row to the Slater determinant with the fucntion $\phi_{\nu}$, and $c_\nu$ removes the first row from the Slater determinant if this row contains $\phi_{\nu}$. Then it's clear why the factor $\left ( -1 \right )^{\sum_{i<\nu}n_i}$ appears: $$\left | n_1,n_2,\dots,n_\nu,\dots,n_N,\dots \right \rangle=\left |\phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu}\dots\phi_{\nu_N}\right \rangle=\left ( -1 \right )^{\sum_{i<\nu}n_i}\left |\phi_{\nu}\phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle$$ But then I don't understand, why a one-particle state of a created or annihilated fermion has to be in the first row of the Slater determinant.
Anyway, I'll accept your answer even if it doesn't use the Slater determinant