We really ought to verify that our implicit definition of the Hodge dual, namely \begin{equation} \alpha \wedge {*}\beta = g(\alpha,\beta)\,\omega \end{equation} is in fact well-defined. To do so, we rely on a standard result about linear maps.

Lemma: Given any linear map $f:V\longmapsto\RR$ on a vector space $V$ with inner product $g$, there is a unique element $\gamma\in V$ such that \begin{equation} f(\alpha) = g(\alpha,\gamma) \end{equation} for every $\alpha\in V$.

The proof of this lemma is obtained by introducing an orthonormal basis $\{\sigma^i\}$, after which one sees that one must have \begin{equation} \gamma = \sum_i\frac{f(\sigma^i)}{g(\sigma^i,\sigma^i)} \>\sigma^i \end{equation}

Since the map \begin{equation} \alpha\longmapsto\alpha\wedge\gamma \end{equation} for $\alpha\in\bigwedge^p$ and fixed $\gamma\in\bigwedge^{n-p}$ is linear (and yields a multiple of the orientation, $\omega$), we can use the lemma to infer the existence of a unique element $h(\gamma)$ such that \begin{equation} \alpha\wedge\gamma = g\bigl(\alpha,h(\gamma)\bigr)\,\omega \end{equation} We can repeat this construction for arbitrary $\gamma$, and it is easily seen that the map $\gamma\longmapsto h(\gamma)$ is itself linear. Furthermore, it is straightforward to evaluate $h$ on our basis. Setting \begin{equation} h(\sigma^J) = a_n \,\sigma^{I_n} \end{equation} where $\{\sigma^{I_m}\}$ is an orthonormal basis of $p$-forms, and $\sigma^J$ is an ($n-p$)-form, we have \begin{equation} \sigma^{I_m}\wedge\sigma^J = g(\sigma^{I_m},a_n\,\sigma^{I_n}) \,\omega = \pm a_m \,\omega \end{equation} Rather than work out the correct sign, it is enough to note that if $J\cap I_m$ is not empty, then $a_m=0$, whereas if the intersection is empty, then $a_m=\pm1$. Thus, $h(\sigma^J)$ must be precisely $\pm\sigma^K$, where $K$ is the dual index set to $J$, so that in particular $\sigma^J\wedge\sigma^K=\pm\omega$.

We are almost there. We have demonstrated the existence of a map $h$ from $p$-forms to $(n-p)$-forms which is almost, but not quite, the Hodge dual. Regardless of signs, the computation above also demonstrates that this map is invertible — and the inverse is the Hodge dual, that is \begin{equation} h({*}\beta) = \beta \end{equation} for $\beta\in\bigwedge^p$. With the wisdom of hindsight, we see that \begin{equation} h(\gamma) = (-1)^{p(n-p)+s} \>{*}\gamma \end{equation} or equivalently, replacing $p$ by $n-p$ (which doesn't change anything), \begin{equation} {*}\beta = (-1)^{p(n-p)+s} \>h(\beta) \end{equation} Since we have established that $h$ is well-defined, so is $*$.