From calculus In manifolds By Spivak,
Poincaré lemma:
Theorem (Poincaré Lemma). If $A \subset \Bbb R^n$ is an open set star-shaped with respect to $0$, then every closed form on $A$ is exact.
proof:
We will define a function $I$ from $l-\text{ forms }$ to $l-1\text{ forms }$ (for each $l$), such that $I(0)=0$ and $\omega=I(dw)+d(Iw)$ for any form $w$. It follows that $\omega=d(Iw)$ if $dw=0.$ Let
$\displaystyle\omega = \sum_{i_1<\ldots<i_l}\omega_{i_1,\ldots i_l}dx^{i_1}\wedge\ldots\wedge dx^{i_l}.$
Since $A$ is star-shaped, we can define
\begin{eqnarray*}Iw(x) &=& \sum_{i_1<\ldots<i_l}\sum_{\alpha=1}^l(-1)^{\alpha-1}\left(\int_0^1t^{l-1}\omega_{i_1,\ldots,i_l}(tx)dt\right)x^{i_\alpha}dx^{i_1}\wedge\ldots\wedge \widehat{dx^{i_\alpha}}\ldots\wedge dx^{i_l}\\ \end{eqnarray*}
(The symbol $\widehat{}$ over $dx^{i_\alpha}$ indicates that it is omitted.)
The proof that $w=Id(w)+d(Iw)$ is an elaborate computation: We have,
\begin{eqnarray*}d(Iw)&=&l.\sum_{i_1<\ldots<i_l}\left(\int_0^1t^{l-1}\omega_{i_1,\ldots,i_l}(tx)dt\right)dx^{i_1}\wedge\ldots\wedge dx^{i_l} + \sum_{i_1<\ldots<i_l}\sum_{\alpha=1}^l\sum_{j=1}^n(-1)^{\alpha-1}\left(\int_0^1t^lD_j(\omega_{i_1,\ldots,i_l})(tx)dt\right)x^{i_\alpha}dx^j\wedge dx^{i_1}\wedge\ldots\wedge \widehat{dx^{i_\alpha}}\ldots\wedge dx^{i_l}\\ \end{eqnarray*}
We also have
\begin{eqnarray*}dw &=&\sum_{i_1<\ldots<i_l}\sum_{j=1}^nD_j(\omega_{i_1,\ldots,i_l})dx^j\wedge dx^{i_1}\wedge\ldots\wedge dx^{i_l}\\ \end{eqnarray*}
Applying $I$ to the $(l+1)-\text{ form }$ $dw$ ,we obtain
\begin{eqnarray*}I(dw) &=& \sum_{i_1<\ldots<i_l}\sum_{j=1}^n\left(\int_0^1t^lD_j(\omega_{i_1,\ldots,i_l})(tx)dt\right)x^jdx^{i_1}\wedge\ldots\wedge dx^{i_l} - \sum_{i_1<\ldots<i_l}\sum_{j=1}^n\sum_{\alpha=1}^l(-1)^{\alpha-1}\left(\int_0^1t^lD_j(\omega_{i_1,\ldots,i_l})(tx)dt\right)x^{i_\alpha}dx^j \wedge dx^{i_1}\wedge\ldots\wedge \widehat{dx^{i_\alpha}}\ldots\wedge dx^{i_l}\\ \end{eqnarray*}
Adding,the triple sums cancel and we obtain
$\begin{aligned} &\color{white}=d(Iw)+I(dw) \\&=\sum_{i_1<\ldots<i_l}l\left(\int_0^1 t^{l-1}\omega_{i_1,\ldots,i_l}(tx)dt\right)dx^{i_1}\wedge\ldots\wedge dx^{i_l} + \sum_{i_1<\ldots<i_l}\sum_{j=1}^n\left(\int_0^1t^lx^jD_j(\omega_{i_1,\ldots,i_l})(tx)dt\right)dx^{i_1}\wedge\ldots\wedge dx^{i_l}\\ &=\sum_{i_1<\ldots<i_l}\int_0^1\frac{d[t^l\omega_{i_1,\ldots,i_l}(tx)]dt}{dt}dx^{i_1}\wedge\ldots\wedge dx^{i_l}\\&=\sum_{i_1<\ldots<i_l}\omega_{i_1,\ldots,i_l}dx^{i_1}\wedge\ldots\wedge dx^{i_l}\\&=\omega\end{aligned}$
Someone can please explain me the step: \begin{eqnarray*}I(dw) &=&\sum_{i_1<\ldots<i_l}\sum_{j=1}^n\left(\int_0^1t^lD_j(\omega_{i_1,\ldots,i_l})(tx)dt\right)x^jdx^{i_1}\wedge\ldots\wedge dx^{i_l} - \sum_{i_1<\ldots<i_l}\sum_{j=1}^n\sum_{\alpha=1}^l(-1)^{\alpha-1}\left(\int_0^1t^lD_j(\omega_{i_1,\ldots,i_l})(tx)dt\right)x^{i_\alpha}dx^j \wedge dx^{i_1}\wedge\ldots\wedge \widehat{dx^{i_\alpha}}\ldots\wedge dx^{i_l}\\ \end{eqnarray*}
I accept both the both should be there here. But why that minus sign is there between them..
The minus sign is there because $dx^j$ is at the front of all these terms and so you effectively add $1$ to all the indices $\alpha-1$ when you apply the definition of $I$.