Generalizing the result $|a\times b|=|a||b||\sin \theta|$ to arbitrary dimensions. The cross product can be generalized to arbitrary dimensions as done below or here. I'm trying to state and prove the general analogue (for arbitrary dimensions) of the equation
$$|a\times b| = |a||b||\sin \theta |$$
where $\theta$ is the angle between $a$ and $b$.

Definition: let $V$ be an inner product space of dimension $n$. The cross product of the vectors $v_1, \ldots ,v_{n-1}$ of $V$ is defined as the unique vector $v$ such that
$$\det (u \ v_1 \ldots v_{n-1}) = u\cdot v$$ for every $u\in V$. It is denoted as $[v_1 \ldots v_{n-1}]$.

Let $v_0, v_1, \ldots , v_{n-1}$ be vectors of an inner product space $V$ (of dimension $n+1$). It is easy to show that the cross product $[v_0, v_1, \ldots ,v_{n-1}]$ is perpendicular to each $v_i$, and has a magnitude equal to the $n$th dimensional "area" of the parallelotope with edges $v_0, v_1, \ldots , v_{n-1}$.
Let $A$ be the $(n-1)$ dimensional area of the parallelotope with edges $v_1, \ldots , v_{n-1}$. That is, let
$$A:=\sqrt{| \det G |}$$
where $G$ is the Gram matrix of the vectors $v_1, \ldots, v_{n-1}$ (i.e. $G_{i,j}=v_i\cdot v_j$). Then the $n$-dimensional area of the parallelotope with edges $v_0, v_1, \ldots , v_{n-1}$ (namely $|[v_0, v_1, \ldots ,v_{n-1}]|$) should be
$$A|v_0||\sin \theta |$$
where $\theta$ is the angle between $v_0$ and its projection $v_0'$ onto the space $\langle v_1, \ldots , v_{n-1} \rangle$, namely
$$v_0':=\sum_{i=1}^{n-1}\frac{v_0\cdot v_i}{|v_i|^2}v_i.$$

Is it true that
$$|[v_0 \ v_1 \ldots v_{n-1}]| = A|v_0||\sin \theta| \ ?$$
If true, how could one prove it? If it isn't true, what am I getting wrong in my attempt to generalize the result stated in the introduction?

My attempt at proving it: the expression $\sin \theta$ can be expanded by using the following definition of angle:
$$\theta := \arccos \left( \frac{v_0\cdot v_0'}{|v_0||v_0'|}\right)$$
together with the fact that $\sin (\arccos (x))=\sqrt{1-x^2}$. Now $A$ can be expanded in multiple ways (i.e. using Leibniz's definition of the determinant or the Laplace expansion). Finally, the $LHS$ can be expanded by noting that
$$[v_0 \ v_1 \ldots v_{n-1}] = \sum_{i=1}^{n+1}(-1)^{i+1}M_{1,i}e_i$$
where $\{ e_1, \ldots , e_{n+1}\}$ is a basis of $V$ and $M_{1,i}$ is the determinant of the submatrix obtained by removing the $i$th row and the $j$th column of the matrix
$$[1 \ v_0 \ v_1 \ldots v_{n-1}] .$$
I've been playing around with the resulting expression trying to force an equality, with no avail.
 A: I hope the following is correct. I have not checked the details. We are working in an $n$-dimensional inner product space.
Definition 1: Cross Product
The cross product, $[v_1,\ldots,v_{n-1}]$, of $n-1$ vectors is the unique vector $v$ such that $\det(u,v_1,\ldots,v_{n-1}) = u\cdot v$.
Definition 2: Volume
The volume of the parallelotope spanned $m$ vectors $v_1,\ldots,v_m$ with $m\leq n$ is $A(v_1,\ldots,v_m) = \sqrt{|\det G|}$ with $G_{ij} = v_i\cdot v_j$.
Lemma 1: If $v_{m+1}$ is orthogonal to the subspace spanned by $v_{1},\ldots,v_m$ then
$$A(v_1,\ldots,v_{m+1}) = A(v_1,\ldots,v_m)|v_{m+1}|$$
Lemma 2: Given $n$ vectors $v_1,\ldots,v_n$ we have
$$A(v_1,\ldots,v_n) = |\det(v_1,\ldots,v_n)|$$
Definition 3: Angles
Given vectors $v_1,\ldots,v_m$ we say that the angle $0\leq \theta \leq \pi/2$ is of $v_m$ with the space spanned by $v_1,\ldots,v_{m-1}$ is determined by
$$\cos(\theta) = \frac{v_{m}\cdot v^\top_{m}}{|v_{m}||v_{m}^\top|}$$
where $v_{m}^\top$ is the orthogonal projection of $v_{m}$ onto the subspace spanned by $v_1,\ldots,v_{m-1}$. We then put,
$$\sin(\theta) = \sqrt{1 - \cos(\theta)^2}$$
Lemma 3: Given vectors $v_1,\ldots,v_m$, and $v_m^\top + v_m^\perp = v_m$ the orthogonal decomposition of $v_m$ with respect to the subspace spanned by $v_1,\ldots,v_{m-1}$, we have
$$|v_{m}^\perp| = |v_{m}|\sin(\theta)$$
Theorem: Given $n-1$ vectors $v_1,\ldots,v_{n-1}$ we have
$$
|[v_1,\ldots,v_{n-1}]| = A(v_1,\ldots,v_{n-2})|v_{n-1}|\sin(\theta)
$$
The proof then becomes a computation, which I show below.
$$|[v_1,\ldots,v_{n-1}]|^2 = \det([v_1,\ldots,v_{n-1}],v_1,\ldots,v_{n-2},v_{n-1}^\perp + v_{n-1}^\top) = \det([v_1,\ldots,v_{n-1}],v_1,\ldots,v_{n-2},v_{n-1}^\perp)$$
first equality is by definition and the second follows by multilinearity of the determinant (and recalling that $v_{n-1}^\top$ is spanned by the other vectors). By Lemma 2 we have
$$\det([v_1,\ldots,v_{n-1}],v_1,\ldots,v_{n-2},v^\perp) = A(v_1,\ldots,v_{n-2},[v_1,\ldots,v_{n-1}],v^\perp)$$
and so
$$|[v_1,\ldots,v_{n-1}]|^2 = A(v_1,\ldots,v_{n-2},[v_1,\ldots,v_{n-1}],v^\perp) = A(v_1,\ldots,v_{n-2})|v^\perp||[v_1,\ldots,v_{n-1}]|$$
where the last equality follows by two successive applications of Lemma 1. The result follows by cancelling and Lemma 3.
