On wikipedia exterior algebra page, section Hodge duality (https://en.wikipedia.org/wiki/Exterior_algebra#Hodge_duality) I read the following definition:
....Then the interior product induces a canonical isomorphism of vector spaces
$ {\textstyle\bigwedge}^k(V^*) \otimes {\textstyle\bigwedge}^n(V) \to {\textstyle\bigwedge}^{n-k}(V) $
by the recursive definition
$ i_{\alpha \wedge \beta} = i_\beta \circ i_\alpha. $
I find the formula natural, but I haven't been able to check it. Furthermore I have not seen anything in the "classical" litterature, where interior product is taken only on a 1-vector.
Can anyone confirm, or tell some references?
You can find it in Werner Greub's Multilinear Algebra, 1st Edition (but not the 2nd Edition).
In short the 1-form interior product you're familar with gives a map $V^* \to \mathrm{End}(\bigwedge V)$ and the universal property of $\bigwedge V^*$ extends it to a homomorphism $\bigwedge V^* \to \mathrm{End}(\bigwedge V)$ (or antihomomorphism in your case, using the opposite algebra of endomorphisms instead).
The isomorphism they reference is called the Poincaré isomorphism and you can find it in Greub also. We choose any nonzero $I \in \bigwedge^nV$ and get (one of) the Poincaré isomorphism(s) $\bigwedge V^* \to \bigwedge V$ via $$ \omega \mapsto i_\omega(I). $$ Another way to construct these isomorphisms is to identify $\bigwedge V^*$ with $(\bigwedge V)^*$ and then note that the bilinear form $\langle\cdot,\cdot\rangle : \bigwedge V\times\bigwedge V \to \mathbb R$ defined by $$ \langle A\wedge B\rangle_n = \langle A,B\rangle I $$ is nondegenerate, where $\langle\cdot\rangle_n$ is the projection onto grade $n$. (We might need to throw in a reversal to get the signs to match up with your convention.) Then $X \mapsto \langle{X,\cdot}\rangle$ is (one of) the Poincaré isomorphism(s) associated with $I$.