The folloing is a problem from Chapter 17 of Lee's Introduction to Smooth Manifolds:
Suppose $U\subseteq \mathbb{R}^n$ is open and star shaped with respect to $0$, and $\omega=\sum'\omega_Idx^I$ is a closed $p$-form on $U$. Show either directly or by tracing through the proof of the Poincaré Lemma, that the $(p-1)$ form $\eta$ given by $$\eta = \sum_I'\sum_{q=1}^p(-1)^{q-1}\left(\int_0^1t^{p-1}\omega_I(tx)dt\right)x^{i_q}dx^{i_1}\wedge...\wedge \widehat{dx^{i_q}}\wedge...\wedge dx^{i_p}$$ satisfies $d\eta =\omega$.
It says to either show it directly, or by tracing through the proof of the Poincaré Lemma. Well, the proof of the Poincaré Lemma in the book simply states that $H(x,t) = c+t(x-c)$ is a homotopy that contracts a domain that's star-shaped with respect to $c$. So I don't see how that helps. So I attempted to show it directly, but I am getting very lost. Here's what I have so far:
I know $$d\left(\left(\int_0^1t^{p-1}\omega_I(tx)dt\right)x^{i_q}dx^{i_1}\wedge...\wedge \widehat{dx^{i_q}}\wedge...\wedge dx^{i_p}\right) \\ = \sum_{j=1}^p\frac{\partial}{\partial x^j}\left(\int_0^1t^{p-1}\omega_I(tx)dtx^{i_q}\right) dx^j\wedge (dx^{i_1}\wedge...\wedge \widehat{dx^{i_q}}\wedge...\wedge dx^{i_p})\\ =\left(\int_0^1t^{p-1}\omega_I(tx)dt \right)dx^I $$ since $\frac{\partial x^{i_q}}{\partial x^j}\neq 0$ only when $j=i_q$, so we get $$d\eta =\sum_I'\sum_{q=1}^p(-1)^{q-1}\int_0^1 t^{p-1}\omega_I(tx)dt dx^I$$ Then I don't know how to go any further.
If $\omega$ were a smooth closed 1-form, I might know how to proceed since there is a Theorem in the book (11.49) which helps with that. The second part of the problem is that it coincides with the potential given in that theorem. Could someone either a) help me finish showing this directly, or b) help me see how the proof of the Poincaré Lemma helps with this problem?