How to see this is a normal vector field? Suppose $M\subseteq\mathbb{R}^{n+1}$ is an $n$-dimensional submanifold, and let $\omega$ be an $n$-form on $M$. Let $\{e_1,...,e_n\}$ be the standard basis of $\mathbb{R}^{n+1}$ and $p$ be the orthogonal projection from the tangent space of $\mathbb{R}^{n+1}$ to the tangent space of $M$ at each point. How does one see that
$$e_1\omega(p(e_2),...,p(e_{n+1}))+e_2\omega(p(e_3),...,p(e_{n+1}),p(e_1))+...+e_{n+1}\omega(p(e_1),...,p(e_n))$$
is a normal vector field on $M$?
 A: The assertion is trivial if $\omega = 0$. So we assume $\omega\neq 0$. By normalizing, we assume that 
$$\tag{1} \omega (v_1, \cdots,v_n) = 1$$ for some orthonormal bassis $\{v_1, \cdots v_n\}$ of the tangent plane. Let $v_{n+1}$ be the unit normal vector. Then 
$$\tag{2} e_j = \sum_{\alpha=1}^{n+1} P_{j\alpha} v_{\alpha}.$$
Note that we have 
$$ p(e_j) = \sum_{\alpha=1}^n P_{j\alpha} v_{\alpha}. $$
This implies
$$\begin{split}
\omega (p(e_{i_1}), \cdots, p(e_{i_n})) &= \sum_{\alpha_1, \cdots, \alpha_n=1}^nP_{i_1\alpha_1} \cdots P_{i_n\alpha_n}\  \omega(v_{\alpha_1}, \cdots, v_{\alpha_n})\\
&= \sum_{\alpha \in S_n}\text{sgn}(\alpha)  P_{i_1\alpha_1} \cdots P_{i_n\alpha_n},
\end{split}$$
where $\text{sgn}(\alpha)$ is the sign of the permutation $(1, \cdots, n)\to (\alpha_1, \cdots, \alpha_n)$. In particular, 
$$\hat \omega_i : =\omega (p(e_{i+1}), \cdots p(e_{n+1}), p(e_1), \cdots p(e_{i-1})) = (-1)^{n+1} (-1)^i \det M_i,$$
where $M_i$ is the minor of $P$ with respect to $(i, n+1)$. So 
$$\begin{split} \omega &:= e_1 \hat\omega_1 + \cdots e_{n+1} \hat\omega_{n+1} \\
&= \sum_{i=1}^{n+1} e_i \hat \omega_i \\
&= \sum_{i=1} ^{n+1} \sum_{\alpha=1}^{n+1} P_{i\alpha} v_\alpha \hat\omega_i \\
&=(-1)^{n+1} \sum_{\alpha=1}^{n+1} \left(\sum_{i=1}^{n+1}(-1)^i P_{i\alpha} \det M_i\right) v_\alpha \\
&= \pm(-1)^{n+1} v_{n+1}. 
\end{split}$$
The last equality holds since for $\alpha \neq n+1$, we have 
$$\tag{3} \sum_{i=1}^{n+1}(-1)^i P_{i\alpha} \det M_i = \det (A_i) = 0,$$
where $A_i$ is the matrix by replacing the $(n+1)$-th column of $P$ by the $\alpha$-th column (thus it has two identical columns). While when $\alpha  =n+1$, the term on the left hand side of $(3)$ equals $\det P$, which is $\pm 1$ since $P$ is orthogonal. 
Thus $\omega $ is up to a sign the unit normal vector $v_{n+1}$. 
