As we know, for $V$ vectoral space and a orientation $\mathcal{O}$ on $V$ and $e_{1},...,e_{n}$, the hodge star operator $\ast:\wedge V^*\rightarrow\wedge V^*$ is defined for

$\ast(e_{1}\wedge...\wedge e_{n})=\pm 1$(acoording to the orientation preserve/invert for the basis $\{e_{i}\}$)

$\ast(1)=\pm e_{1}\wedge...\wedge e_{n}$(again $+$ if $[\{e_{i}\}]=\mathcal{O}$ and $-$ if $[\{e_{i}\}]=-\mathcal{O}$)

$e_{1}\wedge...\wedge e_{k}=\pm e_{k+1}\wedge...\wedge e_{n}$

Already prove that $\ast\ast=(-1)^{n(n-k)}\cdot$ and $<T,S>=\ast(T\wedge\ast S)=\ast(S\wedge \ast T)$ (where $T=\eta_{1}\wedge...\wedge \eta_{1}$, $S=\omega_{1}\wedge...\wedge \omega_{1}$ and $<T,S>=det(<\omega_i,\eta_{j}>)$)

For any $\xi\in V$ denote $\gamma:\ast:\wedge^{k+1} V^*\rightarrow \wedge^{k} V^*$ the adjoint of left exterior multiplication by $\xi$ $$<\gamma(T),S>=<T,\xi\wedge S>$$ for all $T\in\wedge^{k+1} V^*$ and $S\in\wedge^{k} V^*$.

I need to prove that $\gamma(T)=(-1)^{n-k}\ast(\xi\wedge\ast T)$ but I'm a little confused, how prove that $\gamma$ has this expression from this property?