Wedge product of differential and volume form Let $f(x)$ be a $C^1$ function defined on $\mathbb{R}^n$ and $\nabla f(x) \neq 0$ for any $x \in \mathbb{R}^n$. If $d\sigma$ is the volume form on hypersurface $f(x)=c$ induced from $\mathbb{R}^n$ then how to show the equality
$$
    df \wedge d\sigma = |\nabla f| \, dx^1 \wedge \ldots \wedge dx^n?
$$
The main problem is that I don't know to what equal $d\sigma(X_p^1,\ldots,X_p^{n-1})$ for tangent vectors $X_p^1,\ldots,X_p^{n-1} \in T_p \mathbb{R}^n$. Any hint is welcome.
 A: Consider the case of an orientable surface in three-space. The surface has a unit normal ${\bf n} = (n^1,n^2,n^3).$ The volume form of our surface is given by:
$$\operatorname{d}\!\omega = n^1 \operatorname{d}\!y \wedge \operatorname{d}\!z + n^2\operatorname{d}\!z\wedge \operatorname{d}\!x + n^3\operatorname{d}\!x\wedge\operatorname{d}\!y$$
In the case where $M = \{f(x,y,z)=0\}$ the, as you have said,
$${\bf n} = \frac{(f_x,f_y,f_z)}{\sqrt{f_x^2+f_y^2+f_z^2}}$$
It follows that:
$$\operatorname{d}\!\omega = \frac{f_x \operatorname{d}\!y \wedge \operatorname{d}\!z + f_y\operatorname{d}\!z\wedge \operatorname{d}\!x + f_z\operatorname{d}\!x\wedge\operatorname{d}\!y}{\sqrt{f_x^2+f_y^2+f_z^2}}$$
As you already know: $\operatorname{d}\!f = f_z\operatorname{d}\!x + f_y\operatorname{d}\!y +f_z\operatorname{d}\!z$. Hence
\begin{array}{ccc}
\operatorname{d}\!f \wedge \operatorname{d}\!\omega &=& \frac{f_x^2+f_y^2+f_z^2}{\sqrt{f_x^2+f_y^2+f_z^2}}\operatorname{d}\!x\wedge\operatorname{d}\!y\wedge\operatorname{d}\!z \\
\\
&=& |\nabla f| \, \operatorname{d}\!x\wedge\operatorname{d}\!y\wedge\operatorname{d}\!z
\end{array}
