Does every $C^1$ closed differential form differ from some $C^\infty$ closed form by an exact form?

Question: On a $C^\infty$ manifold, does every $C^1$ closed differential form differ from some $C^\infty$ closed form by an exact form?

Motivation: This result holds for $C^1$ closed 1-forms on a $C^\infty$ manifold $M$ for which the first singular homology is finitely generated (e.g., all compact manifolds).

Proof: Let $\omega$ be any $C^1$ closed form on $M$. Let $\sigma_1,\ldots,\sigma_k$ be a $C^\infty$ basis for $H_1(M;\mathbb{R})$ and $[\theta_1],\ldots,[\theta_k] \in H_{dR}^1(M)$ be the dual basis (identifying $H_{dR}^1(M)$ with $Hom(H_1(M;\mathbb{R}),\mathbb{R})$ via the de Rham theorem), where $\theta_1,\ldots,\theta_k$ are $C^\infty$ forms. For $1 \leq i \leq k$, define $c_i := \int_{\sigma_i} \omega$. Then the one-form $\alpha := \omega - c_1 \theta_1 \ldots - c_k \theta_k$ satisfies $\oint_\gamma \alpha = 0$ for any closed loop $\gamma$, since $\gamma$ is a cycle and is thus a finite $\mathbb{R}$-linear combination of the $\sigma_i$ and the integral of $\alpha$ over all $\sigma_i$ is zero.

It follows that given any two points $p,q \in M$, the integral of $\alpha$ over a path from $p$ to $q$ is independent of the path. For any points $p$ and $q$, denote this value by $\int_p^q \alpha$. Fix a basepoint $x_0 \in M$ and define $f:M\to \mathbb{R}$ by $f(p):= \int_{x_0}^p \alpha$. It is easy to show that $df = \alpha$. Hence $\omega - c_1 \theta_1 \ldots,c_k \theta_k = \alpha = df$, showing that $\omega = c_1 \theta_1 \ldots + c_k \theta_k + df$, as desired. This completes the proof.

Note that the proof works if $\omega$ is only differentiable, and in general $f$ is only as smooth as $\omega$.

Observation:

The result would follow for $k$-forms on compact manifolds if it were true that "If the integral of a $C^1$ closed $k$-form $\omega$ over every $k$-cycle was zero, then $\omega$ is exact." If $\omega$ was a $C^\infty$ closed form, then this property could be shown using the de Rham theorem. However, I haven't yet figured out from the de Rham theorem proof whether it would work with $C^1$ forms rather than $C^\infty$ forms.

Yes, this follows, for example, from the fact that we can smooth currents (e.g., differential forms with locally integrable coefficients) and get a chain homotopy formula $$ST - T = d(AT) + A(dT),$$ where $S$ is the smoothing operator. Among other places, you can find this in deRham's classic or Federer's tome.