[proofplan] The result follows immediately from the nilpotence of the [exterior derivative](/theorems/1525), which asserts that $d \circ d = 0$ on every space of smooth differential forms. We unpack the hypothesis of exactness to produce a primitive $\eta$, apply $d$ to the identity $\omega = d\eta$, and invoke nilpotence to conclude $d\omega = 0$. [/proofplan] [step:Unpack the hypothesis of exactness to obtain a primitive] Since $\omega \in \Omega^k(U)$ is exact, by definition there exists a smooth $(k-1)$-form \begin{align*} \eta \in \Omega^{k-1}(U) \end{align*} such that $\omega = d\eta$, where \begin{align*} d : \Omega^{k-1}(U) &\to \Omega^k(U) \end{align*} is the [Exterior Derivative](/theorems/1525). Note that $k \geq 1$ ensures $\Omega^{k-1}(U)$ is well-defined as a space of smooth forms on $U$, so the primitive $\eta$ lives in an admissible degree. [/step] [step:Apply the exterior derivative to both sides of $\omega = d\eta$] Applying the linear operator \begin{align*} d : \Omega^k(U) &\to \Omega^{k+1}(U) \end{align*} to both sides of $\omega = d\eta$ gives \begin{align*} d\omega = d(d\eta) = (d \circ d)\eta. \end{align*} [/step] [step:Invoke nilpotence of the exterior derivative to conclude] By the nilpotence property of the [Exterior Derivative](/theorems/1525), the composition \begin{align*} d \circ d : \Omega^{k-1}(U) &\to \Omega^{k+1}(U) \end{align*} is the zero map; that is, $d(d\alpha) = 0$ for every $\alpha \in \Omega^{k-1}(U)$. Applying this with $\alpha = \eta$ yields \begin{align*} d\omega = d(d\eta) = 0 \quad \text{in } \Omega^{k+1}(U), \end{align*} so $\omega$ is closed. This completes the proof. [guided] The proof is a single application of the fundamental identity $d^2 = 0$, but it is worth recording precisely *why* this identity is the right tool and *what* it depends on. We have $\omega = d\eta$ from the definition of exactness, and we wish to show $d\omega = 0$. The only manipulation available is to apply $d$ to the identity, which gives \begin{align*} d\omega = d(d\eta). \end{align*} This is exactly the input to the nilpotence theorem: for any smooth $(k-1)$-form $\eta$ on an open subset of $\mathbb{R}^n$, the [Exterior Derivative](/theorems/1525) satisfies $d(d\eta) = 0$. At its core, this identity is a coordinate-level restatement of the equality of mixed partial derivatives ($\partial_i \partial_j = \partial_j \partial_i$ acting on smooth functions), combined with the antisymmetry of the wedge product (so symmetric second derivatives are annihilated by antisymmetric wedge sums). Both ingredients require only that $\eta$ have $C^2$ components, which is satisfied because $\eta \in \Omega^{k-1}(U)$ is by definition smooth. We therefore conclude \begin{align*} d\omega = d(d\eta) = 0, \end{align*} which is precisely the statement that $\omega$ is closed. [/guided] [/step]