Let $E,E^*$ be a pair of dual $n$-dimensional vector spaces over a field $\Gamma$ of characteristic zero.
Consider the following result:
Proposition: If $\Delta\ne0$ is an $n$-dimensional determinant function on $E$ (that is, $\Delta:E^n\to\Gamma$ is multilinear and alternating), then for any $(n-1)$-dimensional determinant function $\Phi:E^{n-1}\to\Gamma$ on $E$, there is a unique vector $h\in E$ with $$\Phi(x_1,\ldots,x_{n-1})=\Delta(h,x_1,\ldots,x_{n-1})$$ for all $x_1,\ldots,x_{n-1}\in E$. In other words, $\Phi=i_A(h)\Delta$ where $i_A(h):A^n(E)\to A^{n-1}(E)$ denotes the substitution operator on alternating multilinear functions.
This result explains important properties of determinants like the cofactor expansion, and has a natural geometric interpretation--it tells us for example that measuring oriented area in planes in $\mathbb{R}^3$ is (numerically) equivalent to measuring oriented volume relative to a fixed vector.
I'm interested in finding a more abstract and general statement and proof of this result in the language of exterior algebra. So far I have obtained the following result by considering the isomorphism $A(E)\cong\bigwedge E^*$:
Proposition: If $v^*\in\bigwedge^n E^*$ and $v^*\ne 0$, then for any $u^*\in\bigwedge^{n-1}E^*$ there is a unique vector $h\in E$ with $i(h)v^*=u^*$, where $i(h):\bigwedge^n E^*\to\bigwedge^{n-1}E^*$ is the insertion operator dual to the left multiplication operator $\mu(h)$: $$\langle i(h)v^*,u\rangle=\langle v^*,\mu(h)u\rangle=\langle v^*,h\wedge u\rangle\qquad(u\in\textstyle\bigwedge^{n-1}E)$$ Proof: If $u^*=0$, we take $h=0$. If $u^*\ne 0$, then since $u^*$ has degree $n-1$ it is decomposable and we may write $$u^*=x^*_1\wedge\cdots\wedge x^*_{n-1}\qquad(x^*_i\in E^*)$$ Fix $h^*\in E^*$ with $v^*=h^*\wedge u^*$. Let $F^*$ be the span of $x^*_1,\ldots,x^*_{n-1}$, and let $h$ be the basis vector of $(F^*)^{\perp}$ in $E$ with $$i(h)h^*=\langle h^*,h\rangle=1$$ Since $i(h)$ is an antiderivation with respect to the canonical involution in $\bigwedge E^*$, it follows that $$i(h)v^*=i(h)(h^*\wedge u^*)=i(h)h^*\wedge u^*-h^*\wedge i(h)u^*=u^*$$ The uniqueness of $h$ is clear.
Questions: Is this right? Is there a "better" version?