[proofplan] The proof is the boundaryless special case of the general Stokes theorem. Since $M$ is compact, every smooth form on $M$ has compact support, so the hypotheses of [Stokes' theorem](/theorems/1530) apply to $\omega$. Stokes converts the integral of $d\omega$ over $M$ into the integral of $\omega$ over $\partial M$, and this boundary integral is zero because $\partial M = \varnothing$. [/proofplan] [step:Verify that the form has compact support] Let $\operatorname{supp}\omega \subset M$ denote the closed support of the smooth $(n-1)$-form $\omega$. Since $M$ is compact and $\operatorname{supp}\omega$ is a closed subset of $M$, the set $\operatorname{supp}\omega$ is compact. Hence $\omega$ is compactly supported. [/step] [step:Apply Stokes' theorem and use that the boundary is empty] By the General Stokes Theorem (citing a result not yet in the wiki: General Stokes Theorem), applied to the compactly supported smooth $(n-1)$-form $\omega \in \Omega^{n-1}(M)$ on the oriented smooth $n$-manifold $M$, we have \begin{align*} \int_M d\omega = \int_{\partial M} \omega. \end{align*} Because $M$ is without boundary, $\partial M = \varnothing$. The integral of any form over the empty manifold is zero, so \begin{align*} \int_{\partial M} \omega = \int_{\varnothing} \omega = 0. \end{align*} Therefore \begin{align*} \int_M d\omega = 0. \end{align*} [guided] We want to use Stokes' theorem, whose role is to convert an integral of an [exterior derivative](/theorems/1525) over a manifold into an integral of the original form over the boundary. The form in question is $\omega \in \Omega^{n-1}(M)$, so its exterior derivative $d\omega \in \Omega^n(M)$ is a top-degree form on the oriented smooth $n$-manifold $M$, and therefore $\int_M d\omega$ is defined. The General Stokes Theorem (citing a result not yet in the wiki: General Stokes Theorem) applies to compactly supported smooth $(n-1)$-forms on oriented smooth $n$-manifolds with boundary. We verify the compact-support hypothesis: since $M$ is compact, and since the support $\operatorname{supp}\omega$ is closed in $M$, the set $\operatorname{supp}\omega$ is compact. Thus $\omega$ is compactly supported. Applying Stokes' theorem gives \begin{align*} \int_M d\omega = \int_{\partial M} \omega. \end{align*} Now the hypothesis that $M$ is without boundary is used exactly here: $\partial M = \varnothing$. Integration over the empty manifold gives zero, hence \begin{align*} \int_{\partial M} \omega = \int_{\varnothing} \omega = 0. \end{align*} Combining the two equalities yields \begin{align*} \int_M d\omega = 0. \end{align*} [/guided] [/step]