Let $w=\sum_{I} a_{I}dx_{I}$ be a $k$-form in $\Bbb R ^n$. Let us consider the Hodge operator in a combinatorial form, i.e. as an $(n-k)$ form such that

$$\star(dx_{i_{1}} \wedge \cdots \wedge dx_{i_{k}})=(-1)^{\sigma} (dx_{j_{1}} \wedge \cdots \wedge dx_{j_{n-k}})$$

from where $(i_{1}, \dots, i_{k}, j_{1}, \dots, j_{n-k})$ it is a family of the permutation group, i.e. a permutation of the vector of $n$ entries and $\sigma=0$ if the permutation is even and $\sigma=1$ if the permutation is odd. The idea is to prove the property only with that definition.

Note that: $$ \begin{align} \star \star w&=\star\star \left(\sum_{I} a_{I}dx_{I} \right) \\ &= \star\left(\sum_{I} a_{I} \ast(dx_{i_{1}} \wedge \cdots \wedge dx_{i_{k}}) \right) \\ &=\star \left(\sum_{I} a_{I} \ast(dx_{i_{1}} \wedge \cdots \wedge dx_{i_{k}}) \right) \\ &=\star \left(\sum_{I} a_{I} (-1)^{\sigma_{I}} (dx_{j_{1}} \cdots \wedge dx_{j_{n-k}}) \right) \\ &=\star \left(\sum_{I} a_{I} (-1)^{\sigma_{I}} (dx_{j_{1}} \cdots \wedge dx_{j_{n-k}} \wedge \cdots \wedge dx_{i_{k}}) \right) \\ &=\star \left(\sum_{I} a_{I} (-1)^{\sigma_{I}} (-1)^{k(n-k)} (dx_{i_{1}} \cdots \wedge dx_{i_{k}} \wedge \cdots \wedge dx_{j_{n-k}}) \right) \quad \text{(wedge antisymmetry)}\\ &=(-1)^{k(n-k)} \sum_{I} a_{I} (-1)^{\sigma_{I}} (-1)^{\sigma_{I}} (dx_{j_{1}} \cdots \wedge dx_{j_{n-k}} \wedge \cdots \wedge dx_{i_{k}}) \end{align} $$ From here I think that the result can be deduced, but there is something that does not convince in this proof, note that the permutation is anchored to the counter of $I$, with which, as I can guarantee that when applying once the star comes out $(-1)^{\sigma_{I}}$ it seems to be too convenient, but I can't find the reasoning behind it, any suggestions?

Note: What I want to answer is why when I apply twice the Hodge star, the product $(-1)^{\sigma_{I}} (-1)^{\sigma_{I}}=1$ comes out, because it could happen that when I apply once the star I get the permutation fixed to $I$, i.e. $(\sigma_{I})$ but what guarantees me that when I apply the other star I don't have something like $(-1)^{\sigma_{I}} (-1)^{\sigma_{J}}$