Computing the Hodge-* for a Scalar Let $(E,g)$ be a real oriented inner product space with orthonormal basis $(e_1, \dots, e_n)$ with corresponding
dual basis $(e^1, \dots, e^n)$.  Then, for any $\beta \in \Lambda^0(V) := \mathbb{R}$, how does one prove that
$$
* \beta = \beta \omega_E
$$
where $\omega_E := e^1 \wedge \cdots \wedge e^n$ denotes the induced volume form for $(V,g)$?
I welcome feedback about the correctness of my approach and any alternate approaches that might be more efficient.
 A: First, note that $1 \in \Lambda^0(E)$ is the only positively-oriented orthonormal basis for $\Lambda^0(E)$. 
For, suppose that a nonzero scalar $k$ is basis which we can do since $\Lambda^0(E)$ is one-dimensional. Then if $k$ is
orthonormal with respect to $g$,
$$  
g(k, k) = 1 \implies |k|^2 = 1 \implies k = \pm 1
$$
But, since we require positive orientation this reduces to $k = 1$. So, without loss of generality, 
we take $1$ as a basis for $\Lambda^0(E)$. Also, to be clear, the $g$ referenced here is 
inner product on $\Lambda^k(E)$, $k=1, \dots, n$, that is induced by the inner product on $(V, g)$. Perhaps
a better notation would be $g_0$. 
In any event, the Hodge-$*$ is the unique isomorphism $* \colon \Lambda^0(E) \rightarrow \Lambda^n(E)$ such that
$$
\alpha \wedge (* \beta) = g(\alpha, \beta)\omega_{E}
$$
for all $\alpha \in \Lambda^0(E)$. If follows then from the bilinearity and orthonormality of $g$ that
\begin{align*}
\alpha \wedge (* \beta) &= g(\alpha, \beta)\omega_{E}\\
      &= g(\alpha \cdot 1, \beta \cdot 1) \omega_{E}\\
      &= \alpha \beta g(1,1) \omega_{E}\\
      &= \alpha \beta \omega_{E}.
\end{align*}
But $\alpha$ is just a scalar so
$$
\alpha \wedge (*\beta) = \alpha \cdot (* \beta)
$$
and consequently 
$$
\alpha \cdot (* \beta) = \alpha \beta \omega_E \implies * \beta = \beta \omega_E.
$$
