Given the adjoint of exterior multiplication by $\xi$, that is $\gamma: \Lambda_{p+1}(V) \rightarrow \Lambda_p(V) $, $$\langle \gamma(v), w\rangle = \langle v, \xi\wedge w\rangle$$ for $v\in \Lambda_{p+1}(V)$ and $w\in \lambda_p(V)$. I would like to show that $$\gamma(v) = (-1)^{np} \star(\xi\wedge(\star v))$$ where $\star$ is the Hodge star operator.
I know the basic definition of $\star: \Lambda_p \rightarrow \Lambda_{n-p}$, and given an orthonormal basis $\{e_1,\cdots,e_n\}$ on $V$, we have the computation $$\star (e_{i_1} \wedge e_{i_2}\wedge \cdots \wedge e_{i_k})= e_{i_{k+1}} \wedge e_{i_{k+2}} \wedge \cdots \wedge e_{i_n}.$$ Also for $u,w \in \Lambda_p(V)$, we have $$\star(w\wedge \star v) = \star(v\wedge \star w) = \langle u,w \rangle = \det(u_i, w_j).$$