Algebraic Poincare Lemma In the book "On the De Rham cohomology of Algebraic Varieties" by Robin Hartshorne (http://www.numdam.org/article/PMIHES_1975__45__5_0.pdf), on page 53, Proposition 7.1 states the algebraic version of the Poincare Lemma. It proves that the de Rham complex of the polynomial ring $k[X_1,\ldots,X_n]$ is exact. This is done by induction on $n$.
I cannot really understand the final step. At the end of the page $\omega$ is assumed to be free from $dX_1$ terms. So far it is okay. But next, it is concluded that:
"Since $d\omega=0$ we have $\partial f_{{i_1}\ldots {i_p}} / \partial x_1 =0$ for all $i_1,\ldots, i_p$. Thus the polynomials $f_{{i_1}\ldots {i_p}}$ do not involve $x_1$, at all."
I don't see why we have the partial derivatives become zero and how this concludes the proof.
Thank you very much in advance.
 A: Since $dx_1$ does not appear in $\omega$, the only way to get a $dx_1dx_{i_1}\dots dx_{i_p}$ term in $d\omega$ is from differentiating the term $f_{i_1\dots i_p}dx_{i_1}\dots dx_{i_p}$ in $\omega$ with respect to $x_1$.  That is, the coefficient of $dx_1dx_{i_1}\dots dx_{i_p}$ in $d\omega$ is $\partial f_{{i_1}\ldots {i_p}} / \partial x_1$.  Since $d\omega=0$, this coefficient must be $0$.
As for how this finishes the proof, we are using induction on $n$.  Since all the partial derivatives $\partial f_{{i_1}\ldots {i_p}} / \partial x_1$ are $0$ and $\omega$ does not involve $dx_1$, $\omega$ does not involve the variable $x_1$ at all.  Thus we can instead think of $\omega$ as a $p$-form on the polynomial ring $k[x_2,\dots,x_n]$ with one fewer variable.  By the induction hypothesis, we know the Poincare lemma is valid over that polynomial ring.  That is, there is some $(p-1)$-form $\eta$ over the ring $k[x_2,\dots,x_n]$ such that $d\eta=\omega$, and this equation will still be valid over $k[x_1,\dots,x_n]$.
