Minus Sign in Fermionic Creation and Annihilation Operators 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
 A: 
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.

It doesn't have to be. The order of the rows or columns in a Slater determinant is not physically meaningful; it's only important that you set it up consistently. This is easy to see if you remember that exchanging any two rows or columns of a determinant only changes its sign, not its value. So you could also formulate it so that your creation operator appends a new row at the bottom, and the annihilation operator removes it. Same difference, save for a possible sign change if you compare it to the "top" version. 
With this in mind, the meaning of the $(-1)^{\sum...}$ prefactor can be explained in very simple terms as well; you've already hinted at it in your question. It's the sign change from the successive row swaps that you need to do in order to move a state from somewhere in the middle of the determinant to the position where your operators act on it. Again, whether that be the first or last row, or just any other position in between, is up to you. I assume it's always done to the first row because it makes your manipulations more easy to read when writing the wavefunction in the "linear" form $|\phi_\nu ... \rangle$, because then all the interesting stuff happens at the front of that term instead of its back.
