Let $\alpha$ be a closed (differential) $k$-form with compact support in $\mathbb{R}^{n}$. We want to prove that there exists a $l$-form $\beta$ with compact support in the unit ball of $\mathbb{R}^{n}$ such that $\alpha-\beta$ is exact.
By Poincaré's lemma, we know that
$$H^{k}_{c}(\mathbb{R}^{n})=\begin{cases} 0 &\text{if } k\neq n\\ \mathbb{R} &\text{if } k=n \end{cases}$$
which means that any closed $k$-form with compact support is exact for $k\neq n$. If $k\neq n$, taking $\beta=0$ (which is clearly with compact support in the unit ball) gives $\alpha-\beta=\alpha$ which is exact since $\alpha$ is closed.
So, let $\alpha$ be a closed $n$-form with compact support (I shall denote w.c.s.). We want to construct $\beta$ such that there exists an $(n-1)$-form $\eta$ with compact support verifying $$\alpha-\beta=\text{d}\eta$$
Since
$$H^{n}_{c}(\mathbb{R}^{n})=\frac{\text{Closed $n$-forms w.c.s.}}{\text{Exact $n$-forms w.c.s.}}$$
by definition, it means that for any closed $n$-form $\alpha$ w.c.s. and any exact $n$-form $\text{d}\eta$ w.c.s., we have $(\alpha+\text{d}\eta)\in H^{n}_{c}$
I guess we have to use the fact that $\int(\alpha-\beta)=0$ if and only if $\alpha-\beta$ is exact (here, it is the "only if" part that is of interest). So we have to find $\beta$ such that
$$\text{d}\int\alpha=\beta$$
As $\alpha$ is a $n$ form, we have $\alpha=\sum_{i=1}^{m}f_{i}\text{d}x_{1}\wedge\dots\wedge\text{d}x_{n}$ with $f_{i}\in C^{\infty}_{0}(\mathbb{R}^{n})$ for all $i=1,\dots,m$ but it doesn't lead me anywhere.
I also noticed that $\beta$ has to be closed since $\text{d}^{2}=0$ so that
$$\text{d}(\alpha-\beta)=\text{d}^{2}\eta\iff\text{d}\alpha-\text{d}\beta=0\iff \text{d}\beta=0$$
I do not know if there is something I could do with Stokes'theorem here.
I'm lost... Would you have any indication, please?