The exterior derivative squares to zero

Statement

Let $U \subseteq \mathbb{R}^n$ be open, and let $\omega$ be a differential $k$-form on $U$ with coefficients of class $C^2$. Then $$d(d\omega) = 0.$$

Remarks

Proof: for $\omega = f \, dx_I$, the exterior derivative gives $d(d\omega) = \sum_{j, l} \partial_l \partial_j f \, dx_l \wedge dx_j \wedge dx_I$. By Schwarz's theorem, $\partial_l \partial_j f = \partial_j \partial_l f$, while $dx_l \wedge dx_j = -dx_j \wedge dx_l$. So the terms cancel in pairs, and the terms with $j = l$ are zero. In $\mathbb{R}^3$, the cases $k = 0$ and $k = 1$ are the identities curl of a gradient is zero and divergence of a curl is zero. It follows that each exact form with $C^2$ coefficients is closed.

Explanations

L1 human

Taking the exterior derivative twice always gives zero, because mixed second partial derivatives are equal while the wedge product changes sign. This one rule contains the facts that gradients have no curl and curls have no divergence.