[proofplan] Write the compactly supported $(n-1)$-form in the coordinate basis obtained by omitting one differential at a time. After differentiating, only the derivative in the omitted coordinate can contribute to the top-degree form. The terms omitted in tangential directions integrate to zero by compact support on full real lines, while the term omitted in the normal direction produces exactly the boundary integral, with the sign determined by the induced boundary orientation. [/proofplan]

[step:Expand the form in the standard coordinate basis]For each $j \in \{1,\dots,n\}$, define the coordinate $(n-1)$-form \begin{align*} \theta_j := dx_1 \wedge \cdots \wedge dx_{j-1} \wedge dx_{j+1} \wedge \cdots \wedge dx_n. \end{align*} Since $\{ \theta_1,\dots,\theta_n \}$ is the standard coordinate basis for $(n-1)$-forms on $\mathbb{H}^n$, there exist unique functions \begin{align*} a_j : \mathbb{H}^n \to \mathbb{R}, \qquad a_j \in C_c^\infty(\mathbb{H}^n), \end{align*} such that \begin{align*} \omega = \sum_{j=1}^n a_j \theta_j. \end{align*} Let $K \subset \mathbb{H}^n$ be a compact set containing the supports of all $a_j$. The compact support of $\omega$ implies compact support of each coordinate coefficient because the coefficient functions are obtained by evaluating $\omega$ on the standard coordinate frame. Then there is $R > 0$ such that $K \subset [-R,R]^{n-1} \times [0,R]$. For each $m \in \mathbb{N}$, let $\mathcal{L}^m$ denote $m$-dimensional Lebesgue measure on $\mathbb{R}^m$. In particular, $\mathcal{L}^{n-1}$ is the Lebesgue measure used on the boundary coordinate hyperplane and $\mathcal{L}^1$ is one-dimensional Lebesgue measure.[/step]

[guided]A smooth compactly supported $(n-1)$-form on the coordinate half-space can be decomposed uniquely into the coordinate forms obtained by leaving out one differential. For each index $j \in \{1,\dots,n\}$, set \begin{align*} \theta_j := dx_1 \wedge \cdots \wedge dx_{j-1} \wedge dx_{j+1} \wedge \cdots \wedge dx_n. \end{align*} Thus $\theta_j$ is the form with $dx_j$ omitted. Therefore there are unique coefficient functions \begin{align*} a_j : \mathbb{H}^n \to \mathbb{R}, \qquad a_j \in C_c^\infty(\mathbb{H}^n), \end{align*} such that \begin{align*} \omega = \sum_{j=1}^n a_j \theta_j. \end{align*} The compact support assumption is important because later we integrate one-variable derivatives over either $\mathbb{R}$ or $[0,\infty)$. The compact support of $\omega$ implies compact support of each coordinate coefficient because the coefficient functions are obtained by evaluating $\omega$ on the standard coordinate frame. Let $K \subset \mathbb{H}^n$ be a compact set containing all supports $\operatorname{supp} a_j$. Since $K$ is compact in the closed half-space, there exists $R > 0$ such that \begin{align*} K \subset [-R,R]^{n-1} \times [0,R]. \end{align*} Consequently each coefficient $a_j$ vanishes whenever one of the tangential coordinates has sufficiently large absolute value, or whenever $x_n$ is sufficiently large. For each $m \in \mathbb{N}$, let $\mathcal{L}^m$ denote $m$-dimensional Lebesgue measure on $\mathbb{R}^m$. Thus $\mathcal{L}^{n-1}$ is the Lebesgue measure used on the boundary coordinate hyperplane, and $\mathcal{L}^1$ is the one-dimensional Lebesgue measure used in the normal and tangential one-variable integrations.[/guided]

[step:Compute the exterior derivative of each coordinate component]For each $j \in \{1,\dots,n\}$, \begin{align*} d(a_j \theta_j) = da_j \wedge \theta_j = \sum_{i=1}^n \frac{\partial a_j}{\partial x_i}\, dx_i \wedge \theta_j. \end{align*} If $i \neq j$, then $dx_i$ already occurs in $\theta_j$, so $dx_i \wedge \theta_j = 0$. Hence only $i=j$ remains, and moving $dx_j$ past $dx_1,\dots,dx_{j-1}$ gives \begin{align*} dx_j \wedge \theta_j = (-1)^{j-1} dx_1 \wedge \cdots \wedge dx_n. \end{align*} Therefore \begin{align*} d\omega = \sum_{j=1}^n (-1)^{j-1}\frac{\partial a_j}{\partial x_j}\, dx_1 \wedge \cdots \wedge dx_n. \end{align*}[/step]

[guided]We now compute $d\omega$ term by term. For a fixed $j \in \{1,\dots,n\}$, the form $\theta_j$ has constant coefficients, so $d\theta_j = 0$. The product rule for the [exterior derivative](/theorems/1525) gives \begin{align*} d(a_j \theta_j) = da_j \wedge \theta_j = \sum_{i=1}^n \frac{\partial a_j}{\partial x_i}\, dx_i \wedge \theta_j. \end{align*} Which terms in this sum survive? If $i \neq j$, then $dx_i$ already appears among the factors of $\theta_j$. Since wedging a one-form with itself gives zero, $dx_i \wedge \theta_j = 0$ for all $i \neq j$. Thus the only possible contribution comes from $i=j$: \begin{align*} d(a_j \theta_j) = \frac{\partial a_j}{\partial x_j}\, dx_j \wedge \theta_j. \end{align*} It remains to compute the sign. The form $dx_j \wedge \theta_j$ has $dx_j$ in front of $dx_1,\dots,dx_{j-1}$. To obtain the standard orientation form $dx_1 \wedge \cdots \wedge dx_n$, we move $dx_j$ past exactly $j-1$ one-forms. Each transposition changes the sign, so \begin{align*} dx_j \wedge \theta_j = (-1)^{j-1} dx_1 \wedge \cdots \wedge dx_n. \end{align*} Summing over $j$ gives \begin{align*} d\omega = \sum_{j=1}^n (-1)^{j-1}\frac{\partial a_j}{\partial x_j}\, dx_1 \wedge \cdots \wedge dx_n. \end{align*}[/guided]

[step:Show the tangential derivative terms have zero integral]Fix $j \in \{1,\dots,n-1\}$. By the definition of integration of a top-degree form in the standard orientation, \begin{align*} \int_{\mathbb{H}^n} \frac{\partial a_j}{\partial x_j}\, dx_1 \wedge \cdots \wedge dx_n = \int_{[0,\infty)} \int_{\mathbb{R}^{n-1}} \frac{\partial a_j}{\partial x_j}(x_1,\dots,x_n)\, d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1})\,d\mathcal{L}^1(x_n). \end{align*} For fixed values of all coordinates except $x_j$, the function \begin{align*} t \mapsto a_j(x_1,\dots,x_{j-1},t,x_{j+1},\dots,x_n) \end{align*} is smooth and compactly supported in $\mathbb{R}$. Hence its derivative has integral zero over $\mathbb{R}$. [Fubini's theorem](/theorems/2961) applies because all integrands are smooth with compact support, and therefore \begin{align*} \int_{\mathbb{H}^n} \frac{\partial a_j}{\partial x_j}\, dx_1 \wedge \cdots \wedge dx_n =0. \end{align*} Thus all terms with $j<n$ contribute zero to $\int_{\mathbb{H}^n} d\omega$.[/step]

[guided]We next handle the terms where the omitted differential is tangential to the boundary. Fix $j \in \{1,\dots,n-1\}$. The coordinate $x_j$ ranges over the whole real line, not merely over a half-line. Therefore compact support will force cancellation at both ends. Using the standard orientation on $\mathbb{H}^n$, integration of the top-degree form gives \begin{align*} \int_{\mathbb{H}^n} \frac{\partial a_j}{\partial x_j}\, dx_1 \wedge \cdots \wedge dx_n = \int_{[0,\infty)} \int_{\mathbb{R}^{n-1}} \frac{\partial a_j}{\partial x_j}(x_1,\dots,x_n)\, d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1})\,d\mathcal{L}^1(x_n). \end{align*} The integrand is smooth with compact support, so Fubini's theorem applies. Now hold fixed all coordinates except $x_j$. The resulting one-variable function \begin{align*} t \mapsto a_j(x_1,\dots,x_{j-1},t,x_{j+1},\dots,x_n) \end{align*} is smooth and compactly supported in $\mathbb{R}$. Its derivative integrates to the difference between its limits at $+\infty$ and $-\infty$, both of which are zero because the function has compact support. Hence \begin{align*} \int_{\mathbb{R}} \frac{\partial a_j}{\partial x_j}(x_1,\dots,x_{j-1},t,x_{j+1},\dots,x_n) \, d\mathcal{L}^1(t) =0. \end{align*} Integrating this identity over the remaining coordinates gives \begin{align*} \int_{\mathbb{H}^n} \frac{\partial a_j}{\partial x_j}\, dx_1 \wedge \cdots \wedge dx_n =0. \end{align*} Thus every tangential derivative term, namely every term with $j<n$, contributes zero to the integral of $d\omega$ over $\mathbb{H}^n$.[/guided]

[step:Evaluate the normal derivative term at the boundary]The only remaining contribution is the term with $j=n$: \begin{align*} \int_{\mathbb{H}^n} d\omega = (-1)^{n-1} \int_{\mathbb{H}^n} \frac{\partial a_n}{\partial x_n}\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Using Fubini's theorem and compact support, \begin{align*} \int_{\mathbb{H}^n} \frac{\partial a_n}{\partial x_n}\, dx_1 \wedge \cdots \wedge dx_n &= \int_{\mathbb{R}^{n-1}} \int_{[0,\infty)} \frac{\partial a_n}{\partial x_n}(x_1,\dots,x_{n-1},t) \,d\mathcal{L}^1(t)\,d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}) \\ &= -\int_{\mathbb{R}^{n-1}} a_n(x_1,\dots,x_{n-1},0) \,d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} Therefore \begin{align*} \int_{\mathbb{H}^n} d\omega = (-1)^n \int_{\mathbb{R}^{n-1}} a_n(x_1,\dots,x_{n-1},0) \,d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*}[/step]

[guided]The normal coordinate behaves differently because $x_n$ ranges over $[0,\infty)$ rather than over all of $\mathbb{R}$. From the exterior derivative computation, the $j=n$ term is \begin{align*} (-1)^{n-1}\frac{\partial a_n}{\partial x_n}\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} All other terms have already been shown to integrate to zero, so \begin{align*} \int_{\mathbb{H}^n} d\omega = (-1)^{n-1} \int_{\mathbb{H}^n} \frac{\partial a_n}{\partial x_n}\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Because $\partial a_n/\partial x_n$ is smooth with compact support, Fubini's theorem applies. We integrate first in the normal variable $t=x_n$: \begin{align*} \int_{\mathbb{H}^n} \frac{\partial a_n}{\partial x_n}\, dx_1 \wedge \cdots \wedge dx_n &= \int_{\mathbb{R}^{n-1}} \int_{[0,\infty)} \frac{\partial a_n}{\partial x_n}(x_1,\dots,x_{n-1},t) \,d\mathcal{L}^1(t)\,d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} For fixed $(x_1,\dots,x_{n-1}) \in \mathbb{R}^{n-1}$, the one-variable function \begin{align*} t \mapsto a_n(x_1,\dots,x_{n-1},t) \end{align*} is smooth on $[0,\infty)$ and vanishes for all sufficiently large $t$, because $a_n$ has compact support. Hence \begin{align*} \int_{[0,\infty)} \frac{\partial a_n}{\partial x_n}(x_1,\dots,x_{n-1},t) \,d\mathcal{L}^1(t) = -a_n(x_1,\dots,x_{n-1},0). \end{align*} Substituting this into the iterated integral yields \begin{align*} \int_{\mathbb{H}^n} \frac{\partial a_n}{\partial x_n}\, dx_1 \wedge \cdots \wedge dx_n = -\int_{\mathbb{R}^{n-1}} a_n(x_1,\dots,x_{n-1},0) \,d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} Multiplying by the exterior-derivative sign $(-1)^{n-1}$ gives \begin{align*} \int_{\mathbb{H}^n} d\omega = (-1)^n \int_{\mathbb{R}^{n-1}} a_n(x_1,\dots,x_{n-1},0) \,d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*}[/guided]

[step:Identify the boundary integral with the remaining term]Let \begin{align*} \iota : \partial \mathbb{H}^n &\to \mathbb{H}^n \\ (x_1,\dots,x_{n-1},0) &\mapsto (x_1,\dots,x_{n-1},0) \end{align*} be the inclusion map. Since $\iota^* dx_n = 0$ and $\iota^* dx_i = dx_i$ for $1 \leq i \leq n-1$, all boundary pullbacks $\iota^*(a_j\theta_j)$ vanish except the term $j=n$. Hence \begin{align*} \iota^*\omega = a_n(x_1,\dots,x_{n-1},0)\, dx_1 \wedge \cdots \wedge dx_{n-1}. \end{align*} The induced boundary orientation is represented by \begin{align*} (-1)^n dx_1 \wedge \cdots \wedge dx_{n-1}, \end{align*} so \begin{align*} \int_{\partial \mathbb{H}^n} \omega = (-1)^n \int_{\mathbb{R}^{n-1}} a_n(x_1,\dots,x_{n-1},0) \,d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} This is exactly the value obtained for $\int_{\mathbb{H}^n} d\omega$, and therefore \begin{align*} \int_{\mathbb{H}^n} d\omega = \int_{\partial \mathbb{H}^n} \omega. \end{align*}[/step]

[guided]It remains to match the expression obtained from the normal derivative with the integral of $\omega$ over the boundary. Define the inclusion map \begin{align*} \iota : \partial \mathbb{H}^n &\to \mathbb{H}^n \\ (x_1,\dots,x_{n-1},0) &\mapsto (x_1,\dots,x_{n-1},0). \end{align*} The boundary integral of $\omega$ means the integral of the pullback $\iota^*\omega$ with the induced boundary orientation. We compute the pullback. Since $x_n$ is constantly zero on the boundary, \begin{align*} \iota^* dx_n = 0, \end{align*} while for $1 \leq i \leq n-1$, \begin{align*} \iota^* dx_i = dx_i. \end{align*} If $j<n$, then $\theta_j$ contains $dx_n$, so $\iota^*\theta_j=0$. The only surviving coordinate form is \begin{align*} \theta_n = dx_1 \wedge \cdots \wedge dx_{n-1}. \end{align*} Therefore \begin{align*} \iota^*\omega = a_n(x_1,\dots,x_{n-1},0)\, dx_1 \wedge \cdots \wedge dx_{n-1}. \end{align*} Now we use the boundary orientation convention. The outward normal to $\mathbb{H}^n$ along $\partial \mathbb{H}^n$ is $-\partial_{x_n}$. The boundary orientation is the unique orientation for which placing the outward normal first gives the ambient orientation. This gives the boundary volume form \begin{align*} (-1)^n dx_1 \wedge \cdots \wedge dx_{n-1}. \end{align*} Consequently \begin{align*} \int_{\partial \mathbb{H}^n} \omega = (-1)^n \int_{\mathbb{R}^{n-1}} a_n(x_1,\dots,x_{n-1},0) \,d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} This is the same expression obtained for $\int_{\mathbb{H}^n} d\omega$. Hence \begin{align*} \int_{\mathbb{H}^n} d\omega = \int_{\partial \mathbb{H}^n} \omega. \end{align*}[/guided]