In second quantization, we assume we have fermion operators $a_i$ which satisfy $\{a_i,a_j\}=0$, $\{a_i,a_j^\dagger\}=\delta_{ij}$, $\{a_i^\dagger,a_j^\dagger\}=0$. Another way to say this is that
$$ a_i^\dagger|n_1,...,n_i,...,n_N\rangle = \left\{ \begin{array}{lr} (-1)^{\sum_{j<i} n_j}|n_1,...,n_i+1,...,n_N\rangle & n_i=0\\ 0 &n_i=1 \end{array}\right| $$
$$ a_i|n_1,...,n_i,...,n_N\rangle = \left\{ \begin{array}{lr} (-1)^{\sum_{j<i} n_j}|n_1,...,n_i-1,...,n_N\rangle & n_i=1\\ 0 &n_i=0 \end{array}\right| $$ from which you can derive the relations above.
I understand why the operators on the same sites have to obey the anticommutation relations, since otherwise Pauli exclusion would be violated. I'm not sure I understand why the operators on different sites have to anticommute, however.
Why can't we have an algebra of fermionic operators obeying anticommutation relations for $i=j$, and otherwise obeying the relations $[a_i^{(\dagger)},a_j^{(\dagger)}]=0$? We could define the operators by
$$ a_i^\dagger|n_1,...,n_i,...,n_N\rangle = \left\{ \begin{array}{lr} |n_1,...,n_i+1,...,n_N\rangle & n_i=0\\ 0 &n_i=1 \end{array}\right| $$
$$ a_i|n_1,...,n_i,...,n_N\rangle = \left\{ \begin{array}{lr} |n_1,...,n_i-1,...,n_N\rangle & n_i=1\\ 0 &n_i=0 \end{array}\right| $$ without the sign in front of the ket, from which you can derive the new commutation/anticommutation relations. Is this somehow illegal? Are the operators I've defined not actually well-defined? Is there some way to use the definition I gave to get a contradiction? Or do we just assume the fermion operators anticommute for notational convenience?
So far all the books/pdfs I've looked at prove the anticommutation relations hold for fermion operators on the same site, and then assume anticommutation relations hold on different sites.