This is the problem 16-21 of Lee's Introduction to Smooth manifolds:
Problem 16-21: Let $(M,g)$ be an oriented Riemannian manifold. Show $\iota_X\mathsf{dVol}_g=\star(X^\flat)$ for a vector field $X$.
I know a proof of that as follows: suppose $v_1,v_2,\dots,v_n$ be an orthonormal basis at $p\in M$ then $$\iota_X\mathsf{dVol}_g(v_2,\dots,v_n)=\mathsf{dVol}_g(X,v_2,\dots,v_n)=g(X,v_1)$$ on the other hand and using $X=X^iv_i$ and $X^\flat=g_{ij}X^iE^j$ $$\star(X^\flat)(v_2,\dots,v_n)=\star(g_{ij}X^iE^j)(v_2,\dots,v_n)=g_{ij}X^i\star E^j(v_2,\dots,v_n)=\sum_jg_{ij}X^i \mathsf{dVol}_g(v_2,\dots,v_j,\dots,v_n)=g_{i1}X^i \mathsf{dVol}_g(v_1,\dots,v_n)=g(X,v_1). $$
I messed up a bit in the last two equalities. Anyway I want to know is there another proof and global (basis-free) proof? I think one can prove this just by using properties of interior multiplication and Hodge star. i.e. suppose $\eta$ be an arbitrary $n-1$-form then $$\langle \iota_X\mathsf{dVol}_g,\eta\rangle = \langle \mathsf{dVol}_g,X^\flat\wedge\eta\rangle=??= \langle \star(X^\flat),\eta\rangle $$
How to fill the above gap?
This follows from the definition of the Hodge star operator, where for each one form $\alpha$, $\star \alpha$ is the $(n-1)$-form so that $$\tag{1} \alpha \wedge \star \alpha = \|\alpha\|^2\mathsf{dVol}, $$ here $\|\alpha\|^2 = g^{ij} \alpha_i\alpha_j$.
Then with $\alpha = X^\flat$, $$ \|\alpha\|^2 = \|X\|^2 = g_{ij} X^i X^j$$ and by (1)
$$\| \alpha\|^2 \iota_X \mathsf{dVol}= \iota_X X^\flat\wedge \star X^\flat = \|X\|^2 \star X^\flat.$$
So $ \iota_X \mathsf{dVol} = \star X^\flat$.