Can $e_n$ always be written as a linear combination of $n$-th powers of linear polynomials? User Eric Gregor and I were talking in chat and he mentioned this question and postulated the possibility of an approach through symmetric polynomials. After some thinking, I came to this:

Hypothesis. For any $n$, the elementary symmetric polynomial $e_n\in k[x_1,\cdots,x_n]$ can be expressed as a $k$-linear combination of $n$th powers of degree one homogeneous polynomials.$^\dagger$
$^\dagger$We assume the characteristic does not divide $n!$. We could assume it's zero for more simplicity.

Let us denote $s(a,b,\cdots,c)=(a+b+\cdots+c)^n-(a^n+b^n+\cdots+c^n).$ I have two examples:
$$xy=\frac{s(x,y)}{2} \tag{$n=2$}$$
$$xyz=\frac{s(x,y,z)-\big(s(x,y)+s(y,z)+s(z,x)\big)}{6} \tag{$n=3$}$$
My scratchwork was getting tedious so I didn't finish the $n=4$ case. Besides this I haven't really made any substantive headway, but I did derive the following equality. We denote $\mathrm{pt}\,\lambda$ the number of parts of an integer partition $\lambda\vdash n$, and $m_\lambda$ the sum of all monomials of shape $\lambda$ in $x_j,j\in J$.
$$T_{J,\ell}:=\sum_{\large I\subseteq J \atop \large |I|=\ell}\left(\sum_{i\in I}x_i\right)^n=\sum_{\large \lambda\vdash n \atop \large \mathrm{pt}\lambda\le\ell}\binom{n}{\lambda}\binom{|J|-|I|}{\ell-\mathrm{pt}\,\lambda}m_\lambda.$$
This can be justified as follows: expanding the inner summands of the LHS with the multinomial theorem will result in the terms $m_\lambda$ (with the appropriate multinomial coefficents), times the count of supersets $I$ ($\subseteq J$) of cardinality $\ell$ containing a particular subset $K$ of cardinality $\mathrm{pt}\,\lambda$; construct such $I$ by choosing $|K/I|$ elements out of $|J/I|$ available. In our context $J=[n]$ of course.
The reason I mention this is that the $T_{[n],\ell}$'s appear to be relevant in the computations I was going through for $n=2,3,4$ (as if inverting a linear system in the $m_\lambda$'s...). It may or may not be the correct way of thinking about the problem. I guess my question is then:

*

*Is the hypothesis correct?

*If so, how would we prove it?

*(Optional) If this isn't already inherently answered in the hypothetical proof, how would we explicitly compute what the combinations of powers are?

 A: Yes, the hypothesis is correct. There is an explicit construction. Let
$$
t(i,n) := \sum_{1 \le k_1 < k_2 < \ldots < k_i \le n} s(x_{k_1},\ldots,x_{k_i}).
$$
Then
$$
x_1 \cdots x_n = \frac{1}{n!} \sum_{i=0}^{n-2} (-1)^{i} t(n-i,n).
$$
Proof. Use the definition of $t$ and the multinomial theorem to obtain the equivalent statement
$$
n! \cdot x_1 \cdots x_n = \sum_{i=0}^{n-2} (-1)^i
\sum_{1 \le k_1 < \ldots < k_{n-i} \le n} ~~
\sum_{a_1+\ldots+a_{n-i}=n,~a_j \neq n}
\binom{n}{a_1,\ldots,a_{n-i}} \prod_{j=1}^{n-i} x_{k_j}^{a_j}.
$$
Consider fixed sequences $0< b_1,\ldots,b_m < n$ with $\sum_j b_j = n$ and $1 \le r_1,\ldots,r_m \le n$. The monomial $f := \prod_j x_{r_j}^{b_j}$ appears in all summands of the outer sum with $0 \le i \le n-m$ by choosing $k_j := r_j$ and $a_j := b_j$ for $j \le m$ and $k_j \in (\{1,\ldots,n\} - \{r_j : 1 \le j\le n-i\})$, $b_j := 0$ for $j>m$ (permute indices to meet the "<" requirement).
Obviously there are $\binom{n-m}{n-i-m}$ ways to choose the $k_j$s for fixed $i$. Also observe that the multinomial coefficient is the same for every summand (since joining zeroes doesn't change it).
So the total coefficient of the monomial $f$ on the RHS equals
$ \sum_{i=0}^{n-m} (-1)^i \binom{n-m}{n-m-i} \binom{n}{b_1,\ldots,b_m}$.
For $m<n$, this sum equals $0$. For $n=m$, it equals $n!$, which concludes the proof. 
A: The identity suggested in the accepted answer can also be written without auxiliary polynomials as
$$
 x_1 \cdot \ldots \cdot x_n = \frac{1}{n!} \sum_{I \subseteq \{1,\ldots,n\}} (-1)^{n -|I|} \left(\sum_{i \in I} x_i \right)^n,
$$
where $|I|$ denotes the cardinality of $I$.  Comparison of the coefficients, carried out in the accepted answer, is definitely a legitimate way to check this identity, but one can also use the following argument.
The right-hand side is identically equal to zero for $x_1=0$: Each set $S \subseteq \{2,\ldots,n\}$, not containing $1$, can be paired with the set $\{1\} \cup S$ that contains $1$ and has cardinality $|S|+1$. Consequently, the terms on the right hand-side with $I = S$ and $I = \{1\} \cup S$ cancel out if $x_1 =0$.
This shows that the degree $n$ homogeneous polynomial on the right-hand side is divisible by $x_1$ and, by symmetry, by each of the variables $x_1,\ldots,x_n$. It follows that the polynomial is a constant multiple of $x_1 \cdot \ldots \cdot x_n$. To see that the polynomial is exactly $x_1 \cdot \ldots \cdot x_n$, it suffices to observe that the only source of multiples of the monomial $x_1 \cdot \ldots \cdot x_n$ on the right-hand side is the term for $I = \{1,\ldots,n\}$ and that the coefficient for $x_1 \cdot \ldots \cdot x_n$ in $(x_1 + \cdots +x_n)^n$ is $n!$
