Degree of extension of symmetric function field Let $x_1,\ldots,x_n$ be independent variables.
Let $L=k(x_1,\ldots,x_n)$, the rational function field over $k$ in $x_1,\ldots,x_n$. Consider symmetric polynomials
$$
e_1=x_1+ \cdots + x_n,\\
e_2=\sum_{i<j} x_ix_j,\\
\cdots\\
e_n=x_1x_2\cdots x_n
$$
Let $K=k(e_1,e_2,\ldots,e_n)$.
Claim: $[L:K]=n!$.
A direct way (by some strong results) would solve this immediately. But I am trying to prove with elementary arguments (means, to say precisely, avoiding embeddings of $L$ in $\overline{L}$ which fix $K$, and avoiding fundamental result on elementary symmetric polynomials), because this looks to be a simple example of finite field extensions, which can be introduced in the beginning of course of field theory, and it may be the case that it is easy to prove claim by elementary means.
Initially, I thought about induction on $n$; for $n=2$ this is easier.
For $n=3$, if I want to use case of $n=2$, then we have $[k(x_1,x_2):k(x_1+x_2, x_1x_2)]=2!$.
But $[k(x_1,x_2,x_3):k(x_1,x_2)]$ is infinity, and I confused to proceed in first non-trivial case?
Any hint will be sufficient, I will try to prove. Also, suggest some alternate elementary way for claim, if exists.
 A: Question: "Any hint will be sufficient, I will try to prove. Also, suggest some alternate elementary way for claim, if exists."
Answer: In the case when $G:=S_2$ is the symmetric group on $x_1,x_2$, you get a direct sum decomposition as free $k[s_1,s_2]$-modules
$$A:=k[x_1,x_2]=k[s_1,s_2]\{1\}\oplus k[s_1,s_2]\{v\}:=A^G\oplus A^{-G}$$
where $v:=\frac{1}{2}(x_1-x_2), s_1:=x_1+x_2,s_2:=x_1x_2$. There is an idempotent endomorphism
$$\phi:A\rightarrow A$$
with $\phi^2=\phi$ and where $Im(\phi)=A^G\subseteq A$ is the ring of invariants. There is another idempotent $\psi$ whose image is the submodule $A^{-G}$. By definition $A^{-G}$ is the submodule of elements $x\in A$ with $\sigma(x)=-x$.
The map
$$\psi:A\rightarrow A^{-G}$$
has the property that for any element $f(x_1,x_2)\in A$ it follows
$$\psi(f)=\tilde{f}(x_1,x_2)v,$$
where $\tilde{f}(x_1,x_2)\in A^G$.
From this it follows that $dim_{k(s_1,s_2)}(k(x_1,x_2))=2$.
If $char(k)=0$ there is an operator defined as follows: If $x\in A$ let  $R(x):=\frac{1}{n!}\sum_{g\in S_n}gx.$
You get a map
$$R:A \rightarrow A$$
of left $G$-modules $(G:=S_n)$, and $Im(R)=A^G$. This gives a direct sum decomposition
$$A \cong A^G \oplus Ker(R)$$
of $A$ into $G$-modules. You must study the $A^G$-module $Ker(R)$ and construct a basis as $A^G$-module. What is fun is that you can write down elementary nontrivial examples: The case of $S_3$ can be written out in complete detail. The operator $\psi$ is defined as $\psi(x):=x-R(x)$. In the case of $G:=S_2$ you may write down an explicit formula
$$\psi(x_1^ix_2^j):=\tilde{f}(x_1,x_2)v$$
where $\tilde{f}\in k[s_1,s_2]$ is an explicit $S_2$-invariant polynomial. Similarly for $S_3$.
