Showing equality $\langle F_v\omega, \tau\rangle = \langle \omega, F^*_v\tau\rangle$ 
Let $V$ be an $n$-dimensional vector space with inner product $\langle,\rangle$ and volume element $\mathrm{vol} \in \mathrm{Alt}^n(V)$. Let $v \in \mathrm{Alt}^1(V)$ and $$F_v: \mathrm{Alt}^p(V) \to \mathrm{Alt}^{p+1}(V)$$ be the map $$F_v(\omega)=v\wedge \omega.$$ Show that the map $F^*_v=(-1)^{np} \star \circ F_v \circ \star : \mathrm{Alt}^{p+1}(V) \to \mathrm{Alt}^p(V)$ is adjoint to $F_v$ i.e. $\langle F_v\omega, \tau\rangle = \langle \omega, F^*_v\tau\rangle$.

I'm trying to show this for basis vectors like so. If we pick an orthonormal base $\{e_1, \dots, e_n\}$ for $V$ we get an orthonormal basis $\{\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p} \mid 1 \le i_1 < \dots < i_p \le n\}$ for $\mathrm{Alt}^{p}(V)$. Now I think It would suffice to show that the equality $$\langle F_v(\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}), \varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p} \rangle = \langle \varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}, F^*_v(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p})\rangle$$
holds for two basis vectors $\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}$ and $\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p}$.
Looking at the left-hand side we have $$\begin{align*}\langle F_v(\varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}), \varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p} \rangle &= \langle v \wedge \varepsilon_{i_1} \wedge \dots \wedge \varepsilon_{i_p}, \varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p} \rangle \end{align*}$$
but on the right-hand side I don't know what $F^*_v(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p})$ computes to. I have $$\begin{align*} F^*_v(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p}) &= (-1)^{np} \circ  \star \circ F_v \circ \star(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p}) \end{align*}$$
but I don't have a formula for $\star(\varepsilon_{j_1} \wedge \dots \wedge \varepsilon_{j_p})$. What I do know is that if we instead use shuffles we have $$\star(\varepsilon_{\sigma(1)} \wedge \dots \wedge \varepsilon_{\sigma(p)}) = \mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)}.$$ So this would give $$\begin{align*} F^*_v(\varepsilon_{\sigma(1)} \wedge \dots \wedge \varepsilon_{\sigma(p)}) &= (-1)^{np} \circ  \star \circ F_v \circ \star(\varepsilon_{\sigma(1)} \wedge \dots \wedge \varepsilon_{\sigma(p)}) \\
&= (-1)^{np} \circ  \star \circ F_v(\mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)}) \\ &= (-1)^{np} \circ  \star(v \wedge (\mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)})) \\ &= (-1)^{np} \circ  \star(v) \wedge \star(\mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)})\end{align*}$$
but I'm unfortunatey stuck again as I don't know what $\star(v)$ and $\star(\mathrm{sign}(\sigma) \varepsilon_{\sigma(p+1)} \wedge \dots \wedge \varepsilon_{\sigma(n)})$ evaluates to.
If anyone knows that is this even the right approach I would appreciate the advice?
 A: Let me summarise the discussion in the comments regarding a basis free proof.
The map $\mathbb{R} \to \operatorname{Alt}^nV$ given by $r \mapsto r\operatorname{vol}$ is an isomorphism, so it is enough to show that $\langle F_v(\omega), \tau\rangle\operatorname{vol} = \langle\omega, F_v^*(\tau)\rangle\operatorname{vol}$. Using the fact that $\langle\alpha,\beta\rangle\operatorname{vol} = \alpha\wedge\ast\beta$, we have
\begin{align*}
& \langle\omega, F_v^*(\tau)\rangle\operatorname{vol}\\ 
=&\ \omega\wedge\ast F_v^*(\tau)\\
=&\ \omega\wedge\ast (-1)^{np}\ast F_v(\ast\tau)\\
=&\ \omega\wedge (-1)^{np}\ast\ast F_v(\ast\tau) && (\ast\ \text{is linear})\\
=&\ \omega\wedge (-1)^{np}(-1)^{(n-p)p} F_v(\ast\tau) &&(\ast\ast\ \text{is multiplication by}\ (-1)^{k(n-k)}\ \text{on}\ \operatorname{Alt}^k(V))\\
=&\ \omega\wedge(-1)^{np + np - p^2}F_v(\ast\tau)\\
=&\ \omega\wedge(-1)^pv\wedge\ast\tau &&((-1)^{-p^2} = (-1)^p)\\
=&\ v\wedge\omega\wedge\ast\tau && (\alpha\wedge\beta = (-1)^{ab}\beta\wedge\alpha\ \text{if}\ \alpha\in \operatorname{Alt}^a(V), \beta\in\operatorname{Alt}^b(V))\\
=&\ F_v(\omega)\wedge\ast\tau\\
=&\ \langle F_v(\omega), \tau\rangle\operatorname{vol}.
\end{align*}
