(Combinatorial Nullstellensatz I) Let $f\in \mathbb{F[x_1,x_2\ldots x_n]}$, and let $S_1,S_2\ldots S_n$ be non empty subsets of $\mathbb{F}$. If $f(x)=0$ for all $x\in S_1\times S_2\times \ldots S_n$, then there are polynomials $h_1,h_2\ldots h_n\in \mathbb{F[x_1,x_2\ldots x_n]}$ such that $deg(h_i)\le deg(f)-|S_i|$ and $f(x_1,x_2\ldots x_n)=\sum_\limits{i=1}^{n}h_i(x_1,x_2\ldots x_n)\prod\limits_{s\in S_i}(x_i-s)$.
First please see the following line of proofs. I have a doubt in the last line.
Assume that $g_i(x_i)=\prod\limits_{s\in S_i}(x_i-s)$ for $i=1,2\cdots n$. Now if $x_i\in S_i$ then $g_i(x_i)=0$, that is $x_i^{|S_i|}=\sum\limits_{j=0}^{|S_i|-1}a_{ij}x_i^j$ for $1\le i \le n$.
For $1\le i \le n$, in the polynomial $f$ every occurrences of $x_i^k$ where $k\ge |S_i|$ are replaced by linear combination of smaller power of $x_i$ by using the relation $x_i^{|S_i|}=\sum\limits_{j=0}^{|S_i|-1}a_{ij}x_i^j$. This process will continue until and unless we do not get the power of $x_i$ less than $|S_i|$ for $1\le i \le n$. Let the resultant polynomial be $f'$. Now there exist polynomials $h_i(x_1,x_2\ldots x_n)$ for $1\le i \le n$ such that $f-f'=\sum\limits_{i=1}^{n}h_i(x_1,x_2,\ldots x_n)g_i(x_i)$.
Question: How do I get the polynomials $h_i(x_1,x_2\ldots x_n)$ for $1\le i \le n$?