Proof of Noether's Normalization from Hulek's "Elementary Algebraic Geometry" 
This is Klaus Hulek's Elementary Algebraic geometry. The star-checked part is the thing I can't understand. How can I eliminate all other $x_1, x_1', \ldots ,x_{n-1}, x_{n-1}'$ to make $f$ a polynomial of only $x_n$? 
Thank you all in advance for helping.
 A: The variables $x_1,\dots,x_{n-1}$ are "eliminated" because we have changed to variables $x_1',\dots,x_{n-1}'$. The variables $x_1',\dots,x_{n-1}'$ are not eliminated, per se: it's just that we can view any polynomial in $k[x_1',\dots,x_{n-1}',x_n]$ as an element in $k[x_1',\dots,x_{n-1}'][x_n]$; that is, a polynomial in $x_n$ with coefficients in $k[x_1',\dots,x_n']$. 
When the author writes 

$f(x_1'+\alpha_1 x_n,\dots,x_{n-1}'+\alpha_{n-1}x_n,x_n)=F_d(\alpha_1+\dots+\alpha_{n-1},1)x_n^d+$ terms of lower order in $x_n$.

They mean that $f(x_1'+\alpha_1 x_n,\dots,x_{n-1}'+\alpha_{n-1}x_n,x_n)-F_d(\alpha_1,\dots,\alpha_{n-1},1)x_n^d$ is of lower degree than $d$ as a polynomial in $x_n$ with coefficients in $k[x_1',\dots,x_n']$. 
A: Take $F=F_d$ to be a monomial of the form $X_1^{m_1}...X_n^{m_n}$ with $m_1+\cdots+m_n=d$ then $(X_1+\alpha_1 X_n)^{m_1}\cdots X_n^{m_n}=\alpha_1^{m_1}X_n^{m_1}\alpha_2^{m_2}X_n^{m_2}\cdots X_n^{m_n}$ plus terms of lower-degree in $X_n$, but that's $F(\alpha_1,\ldots,\alpha_{n-1},1)X_n^d$.
Now a general polynomial will have as leading monomial a sum of monomials of the preceding form and you are done.
