[proofplan] The proof is local because equality of differential forms can be checked in coordinate charts. In a chart, write $\omega$ as a finite sum of coefficient functions times wedge products of coordinate one-forms. Applying the coordinate formula for $d$ twice produces second mixed partial derivatives multiplied by antisymmetric wedge products; terms with repeated coordinate one-forms vanish, and the remaining terms cancel in pairs because smooth mixed partial derivatives commute while wedge products change sign. [/proofplan] [step:Reduce the identity to a coordinate chart] Let $(U,\varphi)$ be a smooth coordinate chart on $M$, where $\varphi: U \to \varphi(U) \subseteq \mathbb{R}^n$ is a diffeomorphism onto an open subset of $\mathbb{R}^n$. Let \begin{align*} x_i: U &\to \mathbb{R} \end{align*} denote the $i$-th coordinate function, so $x_i = \pi_i \circ \varphi$, where $\pi_i: \mathbb{R}^n \to \mathbb{R}$ is projection onto the $i$-th coordinate. It suffices to prove \begin{align*} (d(d\omega))|_U = 0 \end{align*} for every coordinate chart $U$, because smooth differential forms are equal on $M$ if their restrictions to every chart domain are equal. [guided] The statement $d(d\omega)=0$ is an equality of smooth $(k+2)$-forms on $M$. Such an equality is local: if a form vanishes after restriction to every chart domain, then it vanishes at every point of $M$, because every point lies in some chart domain. Fix a chart $(U,\varphi)$, with \begin{align*} \varphi: U &\to \varphi(U) \subseteq \mathbb{R}^n. \end{align*} For each index $i \in \{1,\dots,n\}$, define the coordinate function \begin{align*} x_i: U &\to \mathbb{R} \end{align*} by $x_i = \pi_i \circ \varphi$, where $\pi_i$ is projection onto the $i$-th coordinate of $\mathbb{R}^n$. We will prove that the restriction of $d(d\omega)$ to this chart is zero. Since the chart was arbitrary, this proves the global identity. [/guided] [/step] [step:Write the form in the coordinate basis] Let $\mathcal{I}_k$ denote the set of strictly increasing $k$-tuples \begin{align*} I = (i_1,\dots,i_k), \qquad 1 \le i_1 < \cdots < i_k \le n. \end{align*} For $I \in \mathcal{I}_k$, define the coordinate $k$-form \begin{align*} dx_I := dx_{i_1} \wedge \cdots \wedge dx_{i_k}. \end{align*} For $k=0$, take $\mathcal{I}_0$ to consist of the empty tuple and set $dx_\varnothing := 1$. On $U$, the form $\omega$ has a unique expression \begin{align*} \omega|_U = \sum_{I \in \mathcal{I}_k} f_I\, dx_I, \end{align*} where each coefficient is a smooth function \begin{align*} f_I: U &\to \mathbb{R}. \end{align*} [/step] [step:Apply the coordinate formula for the exterior derivative twice] By the [coordinate formula for the exterior derivative](/theorems/3564), \begin{align*} d\omega|_U &= \sum_{I \in \mathcal{I}_k} d f_I \wedge dx_I \\ &= \sum_{I \in \mathcal{I}_k} \sum_{j=1}^n \frac{\partial f_I}{\partial x_j}\, dx_j \wedge dx_I. \end{align*} Applying $d$ again and using that $d(dx_j)=0$ for each coordinate one-form $dx_j$, we get \begin{align*} d(d\omega)|_U &= \sum_{I \in \mathcal{I}_k} \sum_{j=1}^n d\left(\frac{\partial f_I}{\partial x_j}\right) \wedge dx_j \wedge dx_I \\ &= \sum_{I \in \mathcal{I}_k} \sum_{j=1}^n \sum_{\ell=1}^n \frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I. \end{align*} [guided] The coordinate formula for the [exterior derivative](/theorems/1525) says that if a form is written as a sum of smooth coefficients times coordinate wedge forms, then $d$ differentiates the coefficient functions and wedges the resulting one-forms in front. Using \begin{align*} \omega|_U = \sum_{I \in \mathcal{I}_k} f_I\, dx_I, \end{align*} we obtain \begin{align*} d\omega|_U &= \sum_{I \in \mathcal{I}_k} d f_I \wedge dx_I \\ &= \sum_{I \in \mathcal{I}_k} \sum_{j=1}^n \frac{\partial f_I}{\partial x_j}\, dx_j \wedge dx_I. \end{align*} Now apply $d$ once more. Since $d(dx_j)=0$ for coordinate one-forms and the exterior derivative satisfies the graded Leibniz rule, only the coefficient $\frac{\partial f_I}{\partial x_j}$ is differentiated. Thus \begin{align*} d(d\omega)|_U &= \sum_{I \in \mathcal{I}_k} \sum_{j=1}^n d\left(\frac{\partial f_I}{\partial x_j}\right) \wedge dx_j \wedge dx_I \\ &= \sum_{I \in \mathcal{I}_k} \sum_{j=1}^n \sum_{\ell=1}^n \frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I. \end{align*} This is the expression in which the cancellation becomes visible: the coefficient is symmetric in the pair $(j,\ell)$ because the $f_I$ are smooth, while the wedge product is antisymmetric in the same pair. [/guided] [/step] [step:Cancel the terms using symmetry of mixed partials and antisymmetry of the wedge product] Fix $I \in \mathcal{I}_k$. If $j=\ell$, then \begin{align*} dx_\ell \wedge dx_j \wedge dx_I = dx_j \wedge dx_j \wedge dx_I = 0 \end{align*} by alternatingness of the wedge product. If $j \ne \ell$, compare the ordered pairs $(j,\ell)$ and $(\ell,j)$. Since $f_I$ is smooth, the mixed partial derivatives commute: \begin{align*} \frac{\partial^2 f_I}{\partial x_\ell \partial x_j} = \frac{\partial^2 f_I}{\partial x_j \partial x_\ell}. \end{align*} Meanwhile, \begin{align*} dx_\ell \wedge dx_j \wedge dx_I = - dx_j \wedge dx_\ell \wedge dx_I. \end{align*} Therefore the two corresponding summands satisfy \begin{align*} &\frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I + \frac{\partial^2 f_I}{\partial x_j \partial x_\ell}\, dx_j \wedge dx_\ell \wedge dx_I \\ &\qquad = \frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I - \frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I \\ &\qquad = 0. \end{align*} Thus every summand in the double sum either vanishes individually or cancels with its transposed partner, so \begin{align*} d(d\omega)|_U = 0. \end{align*} [guided] We now inspect the terms in \begin{align*} d(d\omega)|_U = \sum_{I \in \mathcal{I}_k} \sum_{j=1}^n \sum_{\ell=1}^n \frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I. \end{align*} There are two cases. First suppose $j=\ell$. Then the wedge product contains the same coordinate one-form twice: \begin{align*} dx_\ell \wedge dx_j \wedge dx_I = dx_j \wedge dx_j \wedge dx_I. \end{align*} Because the wedge product is alternating, $dx_j \wedge dx_j=0$, hence this entire term is zero. Now suppose $j \ne \ell$. The term indexed by $(j,\ell)$ must be paired with the term indexed by $(\ell,j)$. The coefficient functions $f_I$ are smooth, so their second mixed partial derivatives commute: \begin{align*} \frac{\partial^2 f_I}{\partial x_\ell \partial x_j} = \frac{\partial^2 f_I}{\partial x_j \partial x_\ell}. \end{align*} At the same time, the coordinate one-forms anticommute under the wedge product: \begin{align*} dx_\ell \wedge dx_j \wedge dx_I = - dx_j \wedge dx_\ell \wedge dx_I. \end{align*} Therefore the paired terms have equal scalar coefficients and opposite exterior-form factors: \begin{align*} &\frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I + \frac{\partial^2 f_I}{\partial x_j \partial x_\ell}\, dx_j \wedge dx_\ell \wedge dx_I \\ &\qquad = \frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I - \frac{\partial^2 f_I}{\partial x_\ell \partial x_j}\, dx_\ell \wedge dx_j \wedge dx_I \\ &\qquad = 0. \end{align*} Thus the diagonal terms vanish, and the off-diagonal terms cancel in pairs. Hence the full local expression is zero: \begin{align*} d(d\omega)|_U = 0. \end{align*} [/guided] [/step] [step:Conclude the global nilpotence identity] The chart domain $U$ was arbitrary. Since $d(d\omega)$ restricts to zero on every coordinate chart and coordinate charts cover $M$, it follows that \begin{align*} d(d\omega)=0 \end{align*} in $\Omega^{k+2}(M)$. Since $\omega \in \Omega^k(M)$ was arbitrary, the composition \begin{align*} d \circ d : \Omega^k(M) \to \Omega^{k+2}(M) \end{align*} is the zero map. [/step]