Ok, i am trying to prove that $$ \star\star\omega = (-1)^{k(n-k)}\omega$$ if $\omega$ is a $k-form$ in $\mathbb{R}^n $ I only have that $$\star(dx_{i_1}\wedge ... \wedge dx_{i_k})=(-1)^{\sigma}(dx_{j_1}\wedge... \wedge dx_{j_{n-k}})$$ where $(i_1,...,i_k,j_1,...,j_{n-k}$ is a permutation of $(1,...,n)$ and $$i_1 < ... < i_k$$ and $$j_1 < j_{n-k} $$ so $\sigma = 1$ or $0$ depending of the parity of the permutation. Only using this I don't understand how to do it. It is easy to see that applying two times $\star$ it "returns" to $\omega$ but how I get the $(-1)^{j(n-k)}$. I know that there is a way using the Levi-Civita symbol but I can't use it.
Please help with this.
Remember that when you take an orthonormal coframe $\omega_1,\dots,\omega_n$, (e.g., the $dx_i$ when you're in Euclidean space with the usual inner product), then you'll have $$\omega_I \wedge \star\omega_I = \omega_1\wedge\dots\wedge\omega_n\tag1$$ (the volume element). Now, start with $\star\omega_I$. You'll then have to have $$\star\omega_I\wedge\star(\star\omega_I) = \omega_1\wedge\dots\wedge\omega_n\tag2$$ as well. Obviously, as you already noted, $\star(\star\omega_I)=\pm\omega_I$. But to change ($2$) to ($1$) requires switching a $k$ form and an $(n-k)$-form, and such a switch introduces the sign $(-1)^{k(n-k)}$.