The adjoint of left exterior multiplication by $\xi^\flat$ for Hodge star operator As we know, for $V$ a vector space with a symmetric non-degenerate bilinear form $g$ on $V$, an orientation $\mathcal{O}$ on $V$, $e_{1},\dots,e_{n}\in V$ a $g$-orthonormal basis, and $e^1,\dots,e^n$ the associated dual basis on $V^*$, the Hodge star operator $\ast:\Lambda V^*\rightarrow\Lambda V^*$ can be defined as
$\ast(e^{1}\wedge\dots\wedge e^{n})=\pm 1$ ($+$ if $[\{e_{i}\}]=\mathcal{O}$ and $-$ if $[\{e_{i}\}]=-\mathcal{O}$)
$\ast(1)=\pm e^{1}\wedge\dots\wedge e^{n}$ (again $+$ if $[\{e_{i}\}]=\mathcal{O}$ and $-$ if $[\{e_{i}\}]=-\mathcal{O}$)
$*(e^{1}\wedge\dots\wedge e^{k})=\pm e^{k+1}\wedge\dots\wedge e^{n}$
From this it can be proven that $\ast\ast=(-1)^{n(n-k)}\operatorname{id}_{\Lambda V^*}$ and $\langle T,S\rangle=\ast(T\wedge\ast S)=\ast(S\wedge \ast T)$ (where $T=\eta_{1}\wedge\dots\wedge \eta_{k}$, $S=\omega_{1}\wedge\dots\wedge \omega_{k}$ and $\langle T,S\rangle=\det(\langle\omega_i,\eta_{j}\rangle)$).
For any $\xi\in V$ denote $\gamma:\Lambda^{k+1} V^*\rightarrow \Lambda^{k} V^*$ the adjoint of left exterior multiplication by $\xi^\flat=g(-,\xi)$,
$$\langle\gamma(T),S\rangle=\langle T,\xi^\flat\wedge S\rangle,$$
for all $T\in\Lambda^{k+1} V^*$ and $S\in\Lambda^{k} V^*$.
I need to prove that $\gamma(T)=(-1)^{n-k}\ast(\xi^\flat\wedge\ast T)$ but I'm a little confused, how one would prove that $\gamma$ has this expression from this property?
 A: First notice that there exists unique $\gamma:\Lambda^{k+1}V^*\rightarrow \Lambda^k V^*$ such that
$$ \langle \gamma(T),S \rangle = \langle T,\xi\wedge S \rangle .$$
This is due to the non-degeneracy of inner product $ \langle -,- \rangle .$ In fact, assume that there are two such $\gamma_1,\gamma_2.$ Fix an arbitrary $T\in\Lambda^{k+1}V^*.$ Now you get that for every $S\in\Lambda^{k}V^*$
$$ \langle \gamma_1(T),S \rangle = \langle T,\xi\wedge S \rangle = \langle \gamma_2(T),S \rangle .$$
And from non-degeneracy of $ \langle -,- \rangle $ you have that $\gamma_1(T)=\gamma_2(T).$ But $T$ was arbitrary, hence $\gamma_1=\gamma_2.$ Since there is unique $\gamma: \Lambda^ {k+1}V^*\rightarrow  \Lambda^ k V^*$ such that
$$ \langle \gamma(T),S \rangle = \langle T,\xi\wedge S \rangle ,$$
all you need to do is to check whether $\gamma(T)=(-1)^{n-k}\ast(\xi\wedge\ast T)$ has that property. This should be straightforward.
$$ \langle \gamma(T),S \rangle =*(S\wedge *\gamma(T))=*(S\wedge (-1)^{n-k}\ast\ast(\xi\wedge\ast T))=\\=(-1)^{n-k}(-1)^{nk}\ast(S\wedge\xi\wedge\ast T)=(-1)^{n-k}(-1)^{nk}(-1)^{k}\ast(\xi\wedge S\wedge \ast T)=\\=(-1)^{n-k}(-1)^{(n+1)k}\ast((\xi\wedge S)\wedge\ast T)=(-1)^{n(k+1)} \langle T,\xi\wedge S \rangle .$$
Unfortunately I am left with factor $(-1)^{n(k+1)}.$

EDIT
I've found that you have little flaws in your previous calculation. In fact from Warner's book "Foundations of differentiable manifolds and Lie groups," p. 80, we know that

Additionally, in p. 80 too, there is different formula for $\gamma$

Now the calculations go flawless.
$$ \langle \gamma(T),S \rangle =*(S\wedge *\gamma(T))=*(S\wedge (-1)^{nk}\ast\ast(\xi\wedge\ast T))=\\=(-1)^{nk}(-1)^{k(n-k)}\ast(S\wedge\xi\wedge\ast T)=(-1)^{nk}(-1)^{k(n-k)}(-1)^{k}\ast(\xi\wedge S\wedge \ast T)=\\=(-1)^{2nk-k^2+k}\ast((\xi\wedge S)\wedge\ast T)=(-1)^{-k(k-1)} \langle T,\xi\wedge S \rangle = \langle T,\xi\wedge S \rangle .$$
