I'm making a research on Galois theory, and found something interesting regarding the ring of symmetric polynomials: At least up to 5 variables, we can rewrite the elementary symmetric polynomials with powers of symmetric polynomials of degree 1. For the formulation of the question, i will introduce the index set $I_q=\mathbb{N}<q$, where $\mathbb{N}$ are the natural numbers, including zero.
First, we define the subset $S_k\subset I_q^k$ as $$S_k=\{(i_1;...;i_k)\in I_q^k|i_1<i_2<...<i_k\}$$ This allows us to define the $k$-th elementary symmetric polynomial with $q$ variables $e_k=e_k(x_0;...;x_{q-1})$ as: $$e_k=\sum_{(i_1;...;i_k)\in S_k}x_{i_1}...x_{i_k}$$ We also define the symmetric $p$-th powers of first degree polynomials $s_j^p=s_j^p(x_0;...;x_{q-1})$ to be: $$s_j^p=\sum_{(i_1;...;i_j)\in S_k}(x_{i_1}+...+x_{i_j})^p$$ The question is then: can every elementary symmetric polynomial $e_k$ be written as a linear combination of these symmetric $k$-th powers of first degree polynomials $s_j^k$ for an arbitrary number of variables $q$?
As said earlier, i have made some progress, and found a pattern up to $q=5$ variables. I will leave my findings here:
[For $q=1$]
$$e_1=s_1^1$$
[For $q=2$]
$$e_1=s_1^1+0s_2^1$$ $$e_2=\frac{1}{2}(-s_1^2+s_2^2)$$
[For $q=3$]
$$e_1=s_1^1+0s_2^1+0s_3^1$$ $$e_2=\frac{1}{2}(-s_1^2+0s_2^2+s_3^2)$$ $$e_3=\frac{1}{6}(s_1^3-s_2^3+s_3^3)$$ Continuing in a similar fashion, we can construct $q$-dimensional vector spaces $V_k$ over $\mathbb{Q}$, spanned by the set of symmetric $k$-th powers $G_k=\{s_1^k; s_2^k;...;s_q^k\}$. The $k$-th elementary symmetric polynomial is then a vector $\vec{e_k}\in V_k$ in this setting. Note that these representations are not unique, i.e; for $q=2$, $\vec{e_1}=(1;0)=(0;1)$, and hence $G_k$ is not a basis for $V_k$. I'll add a few more values for $q$:
[For $q=4$]
$$\vec{e_1}=(1;0;0;0)$$ $$\vec{e_2}=\frac{1}{2}(-1;0;0;1)$$ $$\vec{e_3}=\frac{1}{6}(4;2;-3;-1)$$ $$\vec{e_4}=\frac{1}{24}(-1;-1;1;1)$$
[For $q=5$]
$$\vec{e_1}=(1;0;0;0;0)$$ $$\vec{e_2}=\frac{1}{2}(-1;0;0;0;1)$$ $$\vec{e_3}=\frac{1}{6}(-6;0;-1;3;0)$$ $$\vec{e_4}=\frac{1}{24}(-5;2;-3;4;-1)$$ $$\vec{e_5}=\frac{1}{120}(1;-1;1;-1;1)$$
etc.