 Solution

Give $\partial X$ the outward-normal-first boundary orientation. The Generalized Stokes theorem states that, for every compactly supported $(n-1)$-form $\omega$,

Choose an oriented coordinate cover by charts into ${\mathbb{R}}^n$ or the half-space ${\mathbb{H}}^n=\{x^1\ge 0\}$, and choose a partition of unity $({\rho }_i)$ subordinate to it. Since the family is locally finite and $\omega$ has compact support, only finitely many ${\rho }_i\omega$ are nonzero. It is therefore legitimate to write both integrals as finite sums and prove the identity for a form supported in one chart.

In an interior chart the integral of an exact compactly supported top form is zero by the fundamental theorem of calculus. In a boundary chart writeIntegrating $d\omega =({\sum }_j{\partial }_j{a}_j)d{x}^1\wedge \cdots \wedge d{x}^n$ coordinate by coordinate kills every tangential derivative. The normal derivative leaves precisely the restriction to $x^1=0$, with the sign selected by the outward-normal-first convention. This is ${\int }_{\partial X}{F}^{\ast }\omega$, proving the theorem.