=== A Special Case === First we suppose that there is a chart $x = \tuple {x_1, \ldots, x_n}: V \subseteq X \to \R^n$ such that $\operatorname{supp} \omega \subseteq V$. {{explain|What is $\operatorname{supp}$}} We may suppose that $V$ is relatively compact. Thus, by composing $x$ with a translation, we may suppose that: :$\displaystyle \map x V \subseteq \mathbb H^n = \set {\tuple {x_1, \ldots, x_n} \in \R^n : x_1 < 0}$ We have, in the coordinates $x$: :$\displaystyle \omega = \sum_{i \mathop = 1}^n f_i \rd x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$ The forms $\hat {\d x}_i := \d x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$ vanish on the tangent space to $\mathbb H^n$ for $i > 1$, so we have: :$(1):\quad \displaystyle \int_{\partial \mathbb H^n} \omega = \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1$ {{explain|Clarification needed of what $\hat {\d x}_i$ means.  The inclusion of $\hat {\d x}_i$ within that long $\wedge$ expression looks wrong to me.}} {{explain|What does $\wedge$ mean in this context?}} Moreover: {{begin-eqn}} {{eqn | l = \d \omega       | r = \sum_{i \mathop = 1}^n \rd f_i \wedge \hat {\d x}_i       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} \d x_i \wedge \hat {\d x}_i       | c = }} {{eqn | r = \paren {\sum_{i \mathop = 1}^n \frac {\partial f} {\partial x_i} } \rd x_1 \wedge \cdots \wedge \d x_n       | c =  }} {{end-eqn}} so that: :$\displaystyle \int_{\mathbb H^n} \rd \omega = \sum_{i \mathop = 1}^n \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots wedge \d x_n$ If $i > 1$: {{begin-eqn}} {{eqn | l = \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots \wedge \d x_n       | r = \int \cdots \int \paren {\int_{-\infty}^\infty \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i       | c = Fubini's Theorem }} {{eqn | r = 0       | c =  }} {{end-eqn}} For $i = 1$: {{begin-eqn}} {{eqn | l = \int_{\mathbb H^n} \frac {\partial f_1} {\partial x_1} \d x_1 \wedge \cdots \wedge \d x_n       | r = \int \cdots \int \paren {\int_{-\infty}^0 \frac {\partial f_i} {\partial x_i} } \hat {\d x}_i       | c = Fubini's Theorem }} {{eqn | r = \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1       | c =  }} {{end-eqn}} So: :$\displaystyle \int_{\mathbb H^n} \rd \omega = \int_{\partial \mathbb H^n} f_1 \, \hat {\d x}_1$ Together with $(1)$, this establishes the result. {{Qed|lemma}} === General Case === Choose a finite family of relatively compact charts $V_1, \ldots, V_k$ on $X$ such that  :$\displaystyle \operatorname {supp} \omega \subseteq \bigcup_{i \mathop = 1}^k V_i$ Choose a partition of unity: :$\chi_1, \ldots, \chi_k$ with $\chi_1 + \cdots + \chi_k = 1$ subordinate to the cover $\set {V_1, \ldots, V_k}$. Put $\omega_i = \chi_i \omega$. Then we have: {{begin-eqn}} {{eqn | l = \omega       | r = \paren {\chi_1 + \cdots + \chi_k} \omega       | c =  }} {{eqn | r = \omega_1 + \cdots + \omega_k       | c =  }} {{end-eqn}} Moreover, $\operatorname {supp} \omega_i \subset V_i$ by definition. Therefore by the special case above, Stokes' theorem holds for each $\omega_i$, so we have: :$\displaystyle \int_X \rd \omega = \sum^k_{i \mathop = 1} \int_x \rd \omega_i = \sum^k_{i \mathop = 1} \int_{\partial X} \omega_i = \int_{\partial X} \omega$ {{Qed}} {{WIP|In progress by BigPansy.}}