[proofplan] The proof is a direct computation in each degree. We expand $d$ on a general $k$-form using the [Coordinate Formula for the Exterior Derivative](/theorems/3564), then collect terms using the antisymmetry of the wedge product. For $k = 0$ we read off the gradient; for $k = 1$ the antisymmetry forces the three classical curl components; for $k = 2$ the three contributions land on the same volume form and sum to the divergence. The vanishing identities $\operatorname{curl} \nabla f = 0$ and $\operatorname{div} \operatorname{curl} F = 0$ follow either from $d \circ d = 0$ or, equivalently, by direct application of the [Symmetry of Second Derivatives](/theorems/332). [/proofplan]

[step:Fix coordinate conventions and recall the exterior derivative formula] We work in the global chart $(U, \operatorname{id})$ with coordinates $x = (x_1, x_2, x_3)$. The spaces $\Omega^k(U)$ of smooth $k$-forms on $U$ admit the global frames \begin{align*} \Omega^0(U) &: \{1\}, \\ \Omega^1(U) &: \{dx_1,\, dx_2,\, dx_3\}, \\ \Omega^2(U) &: \{dx_2 \wedge dx_3,\, dx_3 \wedge dx_1,\, dx_1 \wedge dx_2\}, \\ \Omega^3(U) &: \{dx_1 \wedge dx_2 \wedge dx_3\}. \end{align*} Hence each $\Phi_k$ is a $C^\infty(U)$-linear isomorphism between free $C^\infty(U)$-modules of equal rank. By the [Coordinate Formula for the Exterior Derivative](/theorems/3564), if $\omega = \sum_{|I|=k} \omega_I\, dx_I \in \Omega^k(U)$ with $\omega_I \in C^\infty(U)$ and $dx_I = dx_{i_1} \wedge \cdots \wedge dx_{i_k}$ for $I = (i_1 < \cdots < i_k)$, then \begin{align*} d\omega = \sum_{|I|=k} \sum_{j=1}^{3} \partial_{x_j}\omega_I\, dx_j \wedge dx_I. \end{align*} We use the antisymmetry relations $dx_i \wedge dx_j = -dx_j \wedge dx_i$ and $dx_i \wedge dx_i = 0$. [/step]

[step:Compute $d$ on $\Omega^0(U)$ and identify the gradient] Let $f \in C^\infty(U)$, so $\Phi_0(f) = f \in \Omega^0(U)$. Applying the coordinate formula with $k = 0$, \begin{align*} d(\Phi_0(f)) = df = \sum_{j=1}^{3} \partial_{x_j} f\, dx_j = (\partial_{x_1} f)\, dx_1 + (\partial_{x_2} f)\, dx_2 + (\partial_{x_3} f)\, dx_3. \end{align*} Reading off the coefficients in the frame $\{dx_1, dx_2, dx_3\}$ for $\Omega^1(U)$ and applying $\Phi_1^{-1}$, \begin{align*} \Phi_1^{-1}(df) = (\partial_{x_1} f,\, \partial_{x_2} f,\, \partial_{x_3} f) = \nabla f. \end{align*} Thus $\Phi_1^{-1} \circ d \circ \Phi_0 = \operatorname{grad}$. [/step]

[step:Compute $d$ on $\Omega^1(U)$ and identify the curl]Let $F = (F_1, F_2, F_3) \in C^\infty(U;\mathbb{R}^3)$ and write $\alpha := \Phi_1(F) = \sum_{i=1}^{3} F_i\, dx_i \in \Omega^1(U)$. Applying the coordinate formula, \begin{align*} d\alpha = \sum_{i=1}^{3} \sum_{j=1}^{3} \partial_{x_j} F_i\, dx_j \wedge dx_i. \end{align*} The diagonal terms $j = i$ vanish since $dx_j \wedge dx_j = 0$. Grouping the remaining six terms by the unordered pair $\{i, j\}$ and using $dx_j \wedge dx_i = -dx_i \wedge dx_j$, the coefficient of the ordered basis $2$-form $dx_a \wedge dx_b$ (with $a < b$) is $\partial_{x_a} F_b - \partial_{x_b} F_a$. Concretely, \begin{align*} d\alpha = (\partial_{x_1} F_2 - \partial_{x_2} F_1)\, dx_1 \wedge dx_2 + (\partial_{x_2} F_3 - \partial_{x_3} F_2)\, dx_2 \wedge dx_3 + (\partial_{x_1} F_3 - \partial_{x_3} F_1)\, dx_1 \wedge dx_3. \end{align*} Rewriting the last term using $dx_1 \wedge dx_3 = -dx_3 \wedge dx_1$, \begin{align*} d\alpha = (\partial_{x_2} F_3 - \partial_{x_3} F_2)\, dx_2 \wedge dx_3 + (\partial_{x_3} F_1 - \partial_{x_1} F_3)\, dx_3 \wedge dx_1 + (\partial_{x_1} F_2 - \partial_{x_2} F_1)\, dx_1 \wedge dx_2. \end{align*} Reading off the coefficients in the frame $\{dx_2 \wedge dx_3, dx_3 \wedge dx_1, dx_1 \wedge dx_2\}$ for $\Omega^2(U)$ and applying $\Phi_2^{-1}$, \begin{align*} \Phi_2^{-1}(d\alpha) = \bigl(\partial_{x_2} F_3 - \partial_{x_3} F_2,\; \partial_{x_3} F_1 - \partial_{x_1} F_3,\; \partial_{x_1} F_2 - \partial_{x_2} F_1\bigr) = \operatorname{curl} F. \end{align*} Thus $\Phi_2^{-1} \circ d \circ \Phi_1 = \operatorname{curl}$.[/step]

[guided]We compute $d\alpha$ for $\alpha = \sum_i F_i\, dx_i$. The coordinate formula gives a double sum of $\partial_{x_j} F_i\, dx_j \wedge dx_i$ over $i, j \in \{1, 2, 3\}$. The strategy is to **package this nine-term sum into the three-component curl** by exploiting antisymmetry of $\wedge$. First, the three diagonal terms ($j = i$) vanish because $dx_i \wedge dx_i = 0$. Six terms remain. Pair them by the unordered index set $\{a, b\}$ with $a < b$: each pair contributes \begin{align*} \partial_{x_a} F_b\, dx_a \wedge dx_b + \partial_{x_b} F_a\, dx_b \wedge dx_a = (\partial_{x_a} F_b - \partial_{x_b} F_a)\, dx_a \wedge dx_b, \end{align*} using $dx_b \wedge dx_a = -dx_a \wedge dx_b$. So in the ordered basis $\{dx_1 \wedge dx_2, dx_1 \wedge dx_3, dx_2 \wedge dx_3\}$ of $\Lambda^2$, \begin{align*} d\alpha = (\partial_{x_1} F_2 - \partial_{x_2} F_1)\, dx_1 \wedge dx_2 + (\partial_{x_1} F_3 - \partial_{x_3} F_1)\, dx_1 \wedge dx_3 + (\partial_{x_2} F_3 - \partial_{x_3} F_2)\, dx_2 \wedge dx_3. \end{align*} Now the identification $\Phi_2$ uses the **cyclic** frame $\{dx_2 \wedge dx_3, dx_3 \wedge dx_1, dx_1 \wedge dx_2\}$, not the lexicographic one. Why this frame? Because cycling the indices $(1,2,3) \to (2,3,1) \to (3,1,2)$ keeps the orientation matched to the standard volume form $dx_1 \wedge dx_2 \wedge dx_3$: each cyclic $2$-form $dx_i \wedge dx_j$ satisfies $dx_k \wedge (dx_i \wedge dx_j) = +dx_1 \wedge dx_2 \wedge dx_3$, where $k$ is the missing index. This is precisely the indexing used by the cross product on $\mathbb{R}^3$, which is why we should expect the curl to fall out. To convert, replace $dx_1 \wedge dx_3 = -dx_3 \wedge dx_1$: \begin{align*} d\alpha = (\partial_{x_2} F_3 - \partial_{x_3} F_2)\, dx_2 \wedge dx_3 + (\partial_{x_3} F_1 - \partial_{x_1} F_3)\, dx_3 \wedge dx_1 + (\partial_{x_1} F_2 - \partial_{x_2} F_1)\, dx_1 \wedge dx_2. \end{align*} Reading off the three coefficients in this cyclic frame yields exactly the three components of $\operatorname{curl} F$ as stated.[/guided]

[step:Compute $d$ on $\Omega^2(U)$ and identify the divergence] Let $G = (G_1, G_2, G_3) \in C^\infty(U;\mathbb{R}^3)$ and write \begin{align*} \beta := \Phi_2(G) = G_1\, dx_2 \wedge dx_3 + G_2\, dx_3 \wedge dx_1 + G_3\, dx_1 \wedge dx_2 \in \Omega^2(U). \end{align*} Applying the coordinate formula, \begin{align*} d\beta = dG_1 \wedge (dx_2 \wedge dx_3) + dG_2 \wedge (dx_3 \wedge dx_1) + dG_3 \wedge (dx_1 \wedge dx_2), \end{align*} where $dG_i = \sum_{j=1}^{3} \partial_{x_j} G_i\, dx_j$. In each of the three triple wedges $dx_j \wedge dx_a \wedge dx_b$, the result vanishes unless $\{j, a, b\} = \{1, 2, 3\}$. The surviving contributions are \begin{align*} dG_1 \wedge dx_2 \wedge dx_3 &= \partial_{x_1} G_1\, dx_1 \wedge dx_2 \wedge dx_3, \\ dG_2 \wedge dx_3 \wedge dx_1 &= \partial_{x_2} G_2\, dx_2 \wedge dx_3 \wedge dx_1 = \partial_{x_2} G_2\, dx_1 \wedge dx_2 \wedge dx_3, \\ dG_3 \wedge dx_1 \wedge dx_2 &= \partial_{x_3} G_3\, dx_3 \wedge dx_1 \wedge dx_2 = \partial_{x_3} G_3\, dx_1 \wedge dx_2 \wedge dx_3, \end{align*} where the last two equalities use the cyclic identity $dx_a \wedge dx_b \wedge dx_c = dx_1 \wedge dx_2 \wedge dx_3$ for any even permutation $(a, b, c)$ of $(1, 2, 3)$. Summing, \begin{align*} d\beta = (\partial_{x_1} G_1 + \partial_{x_2} G_2 + \partial_{x_3} G_3)\, dx_1 \wedge dx_2 \wedge dx_3. \end{align*} Applying $\Phi_3^{-1}$, \begin{align*} \Phi_3^{-1}(d\beta) = \partial_{x_1} G_1 + \partial_{x_2} G_2 + \partial_{x_3} G_3 = \operatorname{div} G. \end{align*} Thus $\Phi_3^{-1} \circ d \circ \Phi_2 = \operatorname{div}$. [/step]

[step:Derive the vanishing identities $\operatorname{curl}\nabla f = 0$ and $\operatorname{div}\operatorname{curl} F = 0$]Let $f \in C^\infty(U)$. By Steps 2 and 3, \begin{align*} \Phi_2(\operatorname{curl}(\nabla f)) = d(\Phi_1(\nabla f)) = d(df). \end{align*} We compute $d(df)$ directly: \begin{align*} d(df) = d\!\left(\sum_{i=1}^{3} \partial_{x_i} f\, dx_i\right) = \sum_{i=1}^{3} \sum_{j=1}^{3} \partial_{x_j}\partial_{x_i} f\, dx_j \wedge dx_i. \end{align*} For each unordered pair $\{a, b\}$ with $a \ne b$, the contribution is \begin{align*} \partial_{x_a}\partial_{x_b} f\, dx_a \wedge dx_b + \partial_{x_b}\partial_{x_a} f\, dx_b \wedge dx_a = (\partial_{x_a}\partial_{x_b} f - \partial_{x_b}\partial_{x_a} f)\, dx_a \wedge dx_b. \end{align*} Since $f \in C^\infty(U)$, the [Symmetry of Second Derivatives](/theorems/332) (Schwarz's theorem) gives $\partial_{x_a}\partial_{x_b} f = \partial_{x_b}\partial_{x_a} f$, so each such pair contributes zero. The diagonal pairs $j = i$ vanish via $dx_i \wedge dx_i = 0$. Hence $d(df) = 0$, and since $\Phi_2$ is an isomorphism, $\operatorname{curl}(\nabla f) = 0$. For the second identity, let $F \in C^\infty(U;\mathbb{R}^3)$. By Steps 3 and 4, \begin{align*} \Phi_3(\operatorname{div}(\operatorname{curl} F)) = d(\Phi_2(\operatorname{curl} F)) = d(d(\Phi_1(F))). \end{align*} Writing $\alpha = \Phi_1(F) = \sum_i F_i\, dx_i$ and using the computation of $d\alpha$ from Step 3, \begin{align*} d(d\alpha) = \sum_{i=1}^{3} d(dF_i \wedge dx_i) = \sum_{i=1}^{3}\bigl(d(dF_i) \wedge dx_i - dF_i \wedge d(dx_i)\bigr). \end{align*} By the previous paragraph $d(dF_i) = 0$ for each $i$, and $d(dx_i) = 0$ since $dx_i$ has constant coefficient $1$. Therefore $d(d\alpha) = 0$, and applying $\Phi_3^{-1}$ gives $\operatorname{div}(\operatorname{curl} F) = 0$.[/step]

[guided]Both identities are special cases of $d \circ d = 0$, but we prove them by direct computation to keep the argument self-contained. **Identity 1: $\operatorname{curl}(\nabla f) = 0$.** Apply Step 2 to write $\Phi_1(\nabla f) = df$, then Step 3 to write $\Phi_2(\operatorname{curl}(\nabla f)) = d(df)$. So showing $\operatorname{curl}(\nabla f) = 0$ reduces (since $\Phi_2$ is an isomorphism) to showing $d(df) = 0$. We compute $d(df)$ with $df = \sum_i \partial_{x_i} f\, dx_i$: \begin{align*} d(df) = \sum_{i,j} \partial_{x_j}\partial_{x_i} f\, dx_j \wedge dx_i. \end{align*} Pair the off-diagonal terms by unordered $\{a, b\}$, $a \ne b$: \begin{align*} \partial_{x_a}\partial_{x_b} f\, dx_a \wedge dx_b + \partial_{x_b}\partial_{x_a} f\, dx_b \wedge dx_a = \bigl(\partial_{x_a}\partial_{x_b} f - \partial_{x_b}\partial_{x_a} f\bigr)\, dx_a \wedge dx_b. \end{align*} This is where smoothness enters. The hypothesis $f \in C^\infty(U)$ guarantees in particular that the second partials $\partial_{x_a}\partial_{x_b} f$ and $\partial_{x_b}\partial_{x_a} f$ are continuous, so [Symmetry of Second Derivatives](/theorems/332) applies and the two are equal. Every pair contributes zero. The diagonal $i = j$ contributes zero by antisymmetry $dx_i \wedge dx_i = 0$. Hence $d(df) = 0$, and $\operatorname{curl}(\nabla f) = 0$. **Identity 2: $\operatorname{div}(\operatorname{curl} F) = 0$.** By Steps 3 and 4, $\Phi_3(\operatorname{div}(\operatorname{curl} F)) = d(d\alpha)$ where $\alpha = \Phi_1(F)$. We must show $d(d\alpha) = 0$. Writing $\alpha = \sum_i F_i\, dx_i$, the Leibniz rule for the [exterior derivative](/theorems/1525) on $\Omega^0 \cdot \Omega^1$ products gives, for each $i$, \begin{align*} d(F_i\, dx_i) = dF_i \wedge dx_i + F_i\, d(dx_i) = dF_i \wedge dx_i, \end{align*} since $dx_i$ has the constant coefficient $1$, hence $d(dx_i) = 0$. Now apply $d$ again to $dF_i \wedge dx_i$, using the graded Leibniz rule $d(\eta \wedge \theta) = d\eta \wedge \theta + (-1)^{\deg \eta}\eta \wedge d\theta$: \begin{align*} d(dF_i \wedge dx_i) = d(dF_i) \wedge dx_i - dF_i \wedge d(dx_i) = 0 \wedge dx_i - dF_i \wedge 0 = 0, \end{align*} using $d(dF_i) = 0$ (Identity 1 applied to the scalar function $F_i$) and $d(dx_i) = 0$. Summing over $i$ gives $d(d\alpha) = 0$, and $\operatorname{div}(\operatorname{curl} F) = 0$ follows by applying the isomorphism $\Phi_3^{-1}$. In each case, the **mechanism** behind the vanishing is the same: antisymmetry of $\wedge$ converts a sum of mixed partials into a sum of *differences* of mixed partials, and Schwarz's theorem then forces each difference to vanish. This is the analytic content of $d^2 = 0$ on $\mathbb{R}^n$.[/guided]