I have a doubt regarding the proof that symmetric elementary polynomials are algebrically independent. These are
\begin{aligned}s_{0}(x_{1},x_{2},\dots ,x_{n})&=1,\\[10pt]s_{1}(x_{1},x_{2},\dots ,x_{n})&=\sum _{1\leq j\leq n}x_{j},\\s_{2}(x_{1},x_{2},\dots ,x_{n})&=\sum _{1\leq j<k\leq n}x_{j}x_{k},\\s_{3}(x_{1},x_{2},\dots ,x_{n})&=\sum _{1\leq j<k<l\leq n}x_{j}x_{k}x_{l},\\\end{aligned}
If they are algebrically independent, the polynomials in $\mathbb{K}[x_1 , . . . , x_n]$ can be considered as a polynomial in $s_1 , \ldots , s_n$ which are independent indeterminates. All I have to prove is that $\phi(s_1,\ldots, s_n) \ne 0$.
My manual says that this is true because monomials of different degrees cannot cancel out. For example considering the polynomial $\phi(z_1,\ldots,z_n)$ whose highest degree term is
$az_1^{h_1}z_2^{h_2}\cdots z_n^{h_n}$
and substitute the $\sigma_i$s to yield the following polynomial
$a\sigma_1^{h_1}\sigma_2^{h_2}\cdots \sigma_n^{h_n} = a(x_1+\ldots + x_n)^{h_1}(x_1x_2+\ldots)^{h_2} \cdots (x_1 \cdots x_n)^{h_n}$
whose highest degree term is
$x_1^{h_1+h_2+\ldots+h_n}x_2^{h_2+h_3\ldots+h_n} \cdots x_n^{h_n}$
if $a \ne 0$, this term cannot be canceld out by any other monomial.
My doubt is that even if there is no such monomial able to cancel out the leading term (or any other monoial), why couldn't it be canceled out by the sum (or any other operation) of more monomials?