TITLE: First proof of Poincaré Lemma
QUESTION [15 upvotes]: I know that a way of proving Poincare lemma is to use the homotopy invariance and contractibility of the Euclidean space. Is there is a way of doing it directly (without using the contractibility of $\mathbb{R}^n$)?
What was the first proof of this statement ? I wish to know all the different ways of proving this lemma. Please provide references. Thanks !

REPLY [4 votes]: The above calculation in more detail. Define $h_k:\Omega^p(\mathbb{R}^n)\rightarrow \Omega^{p-1}(\mathbb{R}^n)$ by,
\begin{eqnarray}
h_k(\omega)(x)=\int_0^1t^{p-1}i_X\omega(tx)dt
\end{eqnarray}
where $X=\sum_ix^i\partial_i$, this generates the one parameter group of diffeomorphisms $\Phi_s:x^i\mapsto e^sx^i=:y^i$. To see this note that $X^i(t)=\frac{dx^i(t)}{dt}=x^i(t)$, hence we solve this ODE to find the stated solution. Now, 
\begin{eqnarray}
((h_{k+1}\circ d_k+d_{k-1}\circ h_k)\omega)(x)&=&\int_0^1t^{p-1}\Big(d(i_X\omega)+i_X(d\omega)\Big)(tx)dt\\
&=&\int_0^1t^{p-1}(\mathcal{L}_X\omega)(tx)dt
\end{eqnarray}
Then from the definition of the Lie derivative in terms of the pullback,
\begin{eqnarray}
(\mathcal{L}_X\omega)_x=\Big(\frac{d}{ds}\Big|_{s=0}(\Phi_s)^*\omega\Big)_x
\end{eqnarray}
Now we can compute the pullback explicitly,
\begin{eqnarray}
[((\Phi_s)^*\omega)_x]_{k_1...k_p}=\frac{\partial y^{i_1}}{\partial x^{k_1}}...\frac{\partial y^{i_p}}{\partial x^{k_p}}[\omega_{\Phi_s(x)}]_{i_1...i_p}=e^{ps}[\omega_{e^sx}]_{k_1...k_p}
\end{eqnarray}
Due to linearity we can restrict to the simple case where $\omega(x)=f(x)dx^I$. Therefore, 
\begin{eqnarray}
(\mathcal{L}_X\omega)_x=\Big(\frac{d}{ds}\Big|_{s=0}(\Phi_s)^*\omega\Big)_x=p\omega_x+x^i\frac{\partial f}{\partial x^i}(x)dx^I
\end{eqnarray}
Therefore, 
\begin{eqnarray}
t^{p-1}(\mathcal{L}_X\omega)_{tx}&=&t^{p-1}\Big(p\omega_{tx}+tx^i\frac{\partial f}{\partial x^i}(tx)dx^I\Big)\\
&=&\frac{d}{dt}\Big(t^pf(tx)dx^I\Big)=\frac{d}{dt}(t^p\omega(tx))
\end{eqnarray}
Hence, 
\begin{eqnarray}
((h_{k+1}\circ d_k+d_{k-1}\circ h_k)\omega)(x)=t^p\omega(tx)|_{t=0}^1=\omega(x)
\end{eqnarray}
Therefore since $\omega$ is closed, $d\omega=0$ so $\mathcal{L}_X\omega=d(i_X\omega)$ and hence $d_{k-1}h_k\omega=\omega$, i.e. $\omega$ is exact.