Cross product in $\mathbb R^n$ (from Spivak's book)

Spivak defines cross product in this way:

$\quad$ We conclude this section with a construction which we will restrict to $\mathbf{R}^n$. If $v_1,\ldots,v_{n-1}\in\mathbf{R}^n$ and $\varphi$ is defined by $$\varphi(w)=\det\pmatrix{v_1 \\ \vdots \\ v_{n-1} \\ w},$$ then $\varphi\in\Lambda^1(\mathbf{R}^n)$; therefore there is a unique $z\in\mathbf{R}^n$ such that $$\langle w,z\rangle=\varphi(w)=\det\pmatrix{v_1 \\ \vdots \\ v_{n-1} \\ w}$$ This $z$ is denoted $v_1\times\cdots\times v_{n-1}$ and called the cross product of $v_1,\ldots,v_{n-1}$.

Why such a $z$ exists and why is it unique? When solving problems involving this notion, how do I find this $z$ explicitly (if it's possible)?

Also, what's the meaning of this cross product? Many sources say that the usual cross product in $\mathbb R^3$ can't be generalized to higher dimensions.

Just for kicks, here's a proof of the Riesz representation theorem that Ivo talks about in the (much easier) finite dimension case.

Suppose $V$ is a real, finite dimensional inner product space. Let $f:V\rightarrow \mathbb{R}$ be a linear functional. We claim that there is a unique $z\in V$ for which $f(w) = \langle w,z\rangle$ for all $w\in V$.

First, if $f$ is identically $0$, then $z=0$ works, and only $z=0$ works because $0 =f(z) = \langle z,z\rangle = |z|^2$. Thus, we may assume that $f$ is not identically $0$. Let $v\in V$ with $f(v) \neq 0$.

Linearity implies that $f$ is surjective. In more detail, if $r\in \mathbb{R}$ is any real number, then $f\left(\frac{r}{f(v)} v\right) = \frac{r}{f(v)} f(v) = r$.

By the rank-nullity theorem (which uses finite dimensionality of $V$), it follows that $\ker f\subseteq V$ is a codimension $1$ subspace. Then $(\ker f)^\bot$ is $1$-dimensional. Choose a non-zero $y\in (\ker f)^\bot$. Note that $\langle y,v\rangle \neq 0$, for if it is $0$, then $v\in ((\ker f)^\bot)^\bot = \ker f$, contradicting the fact that $f(v) \neq 0$.

Finally, set $z = \frac{f(v)}{\langle y,v\rangle} y$. I claim that this $z$ works.

For any $w\in V$, we break it up into $w_1 + \lambda v$ where $w_1\in \ker f$.

Then $f(w) = f(\lambda v) = \lambda f(v)$. On the other hand, $$\langle w,z\rangle = \langle w_1 + \lambda v, z\rangle = \lambda \left\langle v, \frac{f(v)}{\langle v,w\rangle}y\right\rangle = \lambda f(v).$$ So they match.

What about uniqueness? If $z_1$ and $z_2$ both work, then for any $w\in V$, we have $\langle w,z_1\rangle = f(w) = \langle w,z_2\rangle$ which implies that $\langle w, z_1 - z_2\rangle = 0$ for all $w\in V$. Choosing $w= z_1 - z_2$, we see that $|z_1 - z_2|^2 = 0$, which means $z_1 = z_2$.

It exists because the inner product $\langle \cdot,\cdot\rangle$ is non-degenerate and we apply the Riesz representation theorem (whose proof in the finite dimensional case is easy) for the linear functional $\varphi\colon \Bbb R^n \to \Bbb R$ given by $$\varphi(w) = \det(v_1,\ldots,v_{n-1},w).$$The Riesz vector $z$ satisfies $\langle w,z\rangle =\varphi(w)$ for all $w \in \Bbb R^n$, and is then denoted by $v_1\times\cdots\times v_{n-1}$. If you write $$v_1\times\cdots \times v_{n-1} = \sum_{i=1}^n \langle e_i,z\rangle e_i$$by orthonormal expansion and compute $\langle e_i, z\rangle = \det(v_1,\cdots,v_{n-1},e_i)$ using Laplace expansion, you'll get a formal expression for $v_1\times\cdots \times v_{n-1}$ with a determinant just like the $3$-dimensional case. For example, in $\Bbb R^4$ we have $$e_1\times e_2 \times e_3 = \begin{vmatrix} e_1 & e_2 & e_3 & e_4 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\end{vmatrix} = e_4,$$and so on.