This is the proof of the following statement made in Complex Manifold and Deformation of Complex Structure Chpt 3, Sec 1 before Thm 3.2
"If $\phi,\psi$ are $C^1$ differential forms over manifold $M$ and $\phi=d\psi$, then $d\phi=0$."(Note that $\psi$ is not $C^2$ form.)
Proof: Due to locality natural, it suffices to prove $d^2\psi=0$. Say $\psi$ is $r-1$ form and $M$ is dimension $n$. Pick $\eta$ a smooth $n-r-1$ form with compact support. $\int_M d^2\psi\wedge\eta=(-1)^{2r-1}\int_M\psi\wedge d^2\eta=0$.(The first equality is true as $\eta$ is compactly supported.) Hence $d^2\psi=0$.
$\textbf{Q:}$ Is $d^2\psi=0$ distribution sense here as this tests over space of $C^1$ differential forms with compact support? Otherwise, I do not see how to conclude $d^2\psi=0$ from $\int_M d^2\psi\wedge\eta=0$ with $\eta$ test differential form. The reason I have trouble with the proof is the notation of $d(d\psi)$'s first differential which makes $d\psi$ as $C^0$ form but this $d\psi$ may not even differentiable.