[proofplan] Write the smooth cycle as a finite smooth singular chain and use the definition of integration over a chain. On each smooth simplex, pull back the exact form and use naturality of the [exterior derivative](/theorems/1525). Stokes theorem converts the integral of $d\beta$ over the chain into the integral of $\beta$ over the boundary chain, and the cycle condition makes that boundary chain zero. [/proofplan] [step:Expand the period over the smooth singular chain] Write the smooth singular $k$-chain $c$ as \begin{align*} c = \sum_{j=1}^N a_j \sigma_j, \end{align*} where $N \in \mathbb{N}$, each coefficient $a_j \in \mathbb{R}$, and each simplex is a smooth map \begin{align*} \sigma_j : \Delta^k \to M \end{align*} from the standard oriented $k$-simplex $\Delta^k$ to $M$. Since $c$ is a cycle, its boundary satisfies \begin{align*} \partial c = \sum_{j=1}^N a_j \partial \sigma_j = 0. \end{align*} By definition of integration of a differential form over a smooth singular chain, \begin{align*} \int_c \alpha = \sum_{j=1}^N a_j \int_{\Delta^k} \sigma_j^*\alpha. \end{align*} [/step] [step:Pull exactness back to each simplex] Since $\alpha = d\beta$ on $M$, the naturality of the exterior derivative under pullback gives, for each $j \in \{1,\dots,N\}$, \begin{align*} \sigma_j^*\alpha = \sigma_j^*(d\beta) = d(\sigma_j^*\beta). \end{align*} Therefore \begin{align*} \int_c \alpha = \sum_{j=1}^N a_j \int_{\Delta^k} d(\sigma_j^*\beta). \end{align*} [/step] [step:Apply Stokes theorem on each smooth simplex] For each $j$, the form $\sigma_j^*\beta$ is a smooth $(k-1)$-form on the oriented simplex $\Delta^k$. By Stokes theorem applied to $\Delta^k$ with its oriented boundary (citing a result not yet in the wiki: Stokes Theorem), \begin{align*} \int_{\Delta^k} d(\sigma_j^*\beta) = \int_{\partial \Delta^k} \sigma_j^*\beta. \end{align*} Thus \begin{align*} \int_c \alpha = \sum_{j=1}^N a_j \int_{\partial \Delta^k} \sigma_j^*\beta = \int_{\partial c} \beta. \end{align*} [guided] The point of this step is to move from an integral over the $k$-dimensional chain to an integral over its oriented boundary. For each $j$, the pullback $\sigma_j^*\beta$ is a smooth $(k-1)$-form on the oriented simplex $\Delta^k$, so Stokes theorem applies to the oriented manifold with boundary $\Delta^k$ and gives \begin{align*} \int_{\Delta^k} d(\sigma_j^*\beta) = \int_{\partial \Delta^k} \sigma_j^*\beta. \end{align*} The right-hand side is exactly the contribution of the boundary simplex $\partial\sigma_j$ to the chain integral of $\beta$. Multiplying by $a_j$ and summing over $j$ gives \begin{align*} \sum_{j=1}^N a_j \int_{\Delta^k} d(\sigma_j^*\beta) = \sum_{j=1}^N a_j \int_{\partial\sigma_j} \beta = \int_{\partial c} \beta. \end{align*} This is the chain-level form of Stokes theorem: \begin{align*} \int_c d\beta = \int_{\partial c} \beta. \end{align*} [/guided] [/step] [step:Use the cycle condition to make the boundary integral vanish] Because $c$ is a cycle, $\partial c = 0$. Integration over the zero chain is zero by linearity of chain integration, hence \begin{align*} \int_{\partial c} \beta = \int_0 \beta = 0. \end{align*} Combining this with the previous step yields \begin{align*} \int_c \alpha = 0. \end{align*} This proves that every exact $k$-form has zero period over every smooth oriented $k$-cycle. [/step]