Hodge star operator and exterior calculation

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I am learning complex geometry by D. Huybrechts. Here is a formula that I can't understand $$\omega \wedge \beta\wedge \star \alpha=\beta\wedge(\omega \wedge \star \alpha)\tag 1$$

Here $\omega$ is the fundamental form which is a $2$-form actually. And I try to expand both sides by the definition $\alpha \wedge \star \beta= \langle \alpha, \beta \rangle \cdot vol$

For LHS, we have $\langle \omega \wedge \beta,\alpha \rangle \cdot vol$ and for RHS, we have $\beta \wedge \langle \omega, \alpha \rangle \cdot vol$ But here is the things I don't understand:

(1) And let's suppose $\alpha,\beta \in \land^k V^{*}$. For $\langle \omega \wedge \beta,\alpha \rangle$, here $\omega \wedge \beta$ is a $(k+2)$- form so how can an inner product operated by a $k+2$-form and a $k$-form ? So does for RHS.

(2) How to prove $(1)$ by expanding in my way or the other methods?

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As $\omega$ is a $2$-form, it commutes for the $\wedge$-product with other $p$-forms, (because if $\alpha$ and $\beta$ are $p$ and $q$-forms, $\alpha\wedge \beta = (-1)^{pq} \beta\wedge \alpha$). Moreover, this product is associative. Forget about the $\star$ in $(\star \alpha)$ and consider only $(\star\alpha)$ as a differential form. Then : \begin{align} \omega \wedge \beta \wedge \star \alpha &= \left(\omega \wedge \beta\right) \wedge \star \alpha \\ &= \left(\beta \wedge \omega \right)\wedge \star \alpha \\ &= \beta \wedge \left( \omega \wedge \star \alpha\right) \end{align}