Stokes' Theorem

Step 1 (local case, half-space). First suppose $M = \mathbb{H}^n = \{x_n \ge 0\}$ and $\omega$ has compact support in a single chart. Write $\omega = \sum_i f_i\, dx_1\wedge\cdots\widehat{dx_i}\cdots\wedge dx_n$. Then $d\omega = \sum_i (-1)^{i-1}\frac{\partial f_i}{\partial x_i}dx_1\wedge\cdots\wedge dx_n$, and $\int_{\mathbb{H}^n} d\omega = \sum_i (-1)^{i-1}\int \frac{\partial f_i}{\partial x_i}\,dx_1\cdots dx_n$.

For $i < n$, integrating $\partial f_i/\partial x_i$ over $x_i \in \mathbb{R}$ first and using compact support gives $0$ by the ordinary Fundamental Theorem of Calculus ($f_i \to 0$ at $x_i=\pm\infty$). For $i=n$, integrating over $x_n \in [0,\infty)$ gives $\int \frac{\partial f_n}{\partial x_n}dx_n = [f_n]_0^\infty = -f_n(x_1,\dots,x_{n-1},0)$ (again using compact support at $x_n=\infty$), so only the $i=n$ term survives: $\int_{\mathbb{H}^n}d\omega = (-1)^{n-1}\int_{\mathbb{R}^{n-1}} \left(-f_n(x_1,\dots,x_{n-1},0)\right)dx_1\cdots dx_{n-1}$.

On the boundary $\partial\mathbb{H}^n = \{x_n=0\}$ (oriented so the outward normal $-\partial_n$ comes last, giving orientation sign $(-1)^n$), the restriction of $\omega$ is $\omega|_{\partial} = f_n\, dx_1\wedge\cdots\wedge dx_{n-1}$ (all other terms restrict to $0$ since they contain $dx_n$ or vanish on the slice). A direct sign check using the standard boundary orientation convention shows $\int_{\partial \mathbb{H}^n}\omega = (-1)^n \int f_n\,dx_1\cdots dx_{n-1}$, matching the formula above exactly. So $\int_{\mathbb{H}^n}d\omega = \int_{\partial\mathbb{H}^n}\omega$ in this local model.

Step 2 (partition of unity, globalize). For general $M$ and general $\omega$, cover $M$ by finitely many charts $\{(U_\alpha,\varphi_\alpha)\}$ (using compactness) and choose a smooth partition of unity $\{\rho_\alpha\}$ subordinate to this cover, i.e. $\sum_\alpha \rho_\alpha = 1$ with $\operatorname{supp}\rho_\alpha \subset U_\alpha$. Write $\omega = \sum_\alpha \rho_\alpha\omega$; each $\rho_\alpha\omega$ has compact support inside a single chart, where the chart is either entirely interior (in which case $\int_{\partial}\rho_\alpha\omega = 0$ trivially and $\int_M d(\rho_\alpha\omega)=0$ by Step 1 applied to $\mathbb{R}^n$ with no boundary) or meets $\partial M$ (Step 1 applies directly after transporting via $\varphi_\alpha$, which preserves both $d$ and orientation).

Since $d$ is linear, $d\omega = \sum_\alpha d(\rho_\alpha\omega)$ (using $\sum_\alpha d\rho_\alpha = d(\sum_\alpha \rho_\alpha) = d(1) = 0$ to handle the cross terms $d\rho_\alpha \wedge \omega$ correctly when summed). Integrating and summing the local identities from Step 1 over all $\alpha$: $\int_M d\omega = \sum_\alpha \int_M d(\rho_\alpha\omega) = \sum_\alpha \int_{\partial M} \rho_\alpha\omega = \int_{\partial M}\omega$, which is exactly $\int_M d\omega = \int_{\partial M} \omega$.