Product of $k$ factors as linear combination of linear combinations of its factors that are raised to the power of $k$

For $k=4$ we have

$$x_1 x_2 x_3 x_4 = \frac{1}{192} ({x_1}+{x_2}+{x_3}+{x_4})^4+\frac{1}{384} \left(({x_1}+{x_2}-{x_3}-{x_4})^4+({x_1}-{x_2}+{x_3}-{x_4})^4+(-{x_1}+{x_2}+{x_3}-{x_4})^4+({x_1}-{x_2}-{x_3}+{x_4})^4+(-{x_1}+{x_2}-{x_3}+{x_4})^4+(-{x_1}-{x_2}+{x_3}+{x_4})^4\right)+\frac{1}{192} \left(-({x_1}+{x_2}+{x_3}-{x_4})^4-({x_1}+{x_2}-{x_3}+{x_4})^4-({x_1}-{x_2}+{x_3}+{x_4})^4-(-{x_1}+{x_2}+{x_3}+{x_4})^4\right) $$

Maybe you recognize a pattern, the numbers in the denominator are for instance $192=4 \cdot 6 \cdot 8$ and $384=2 \cdot 4 \cdot 6 \cdot 8$.

They can be found like so:

• Construct an approach that is fully symmetric in all variables $x_i$ and that involves linear parameters $a,b,c$.
• Expand and collect w.r.t. $a,b,c$
• Compare coefficients. E.g. in the above case, replacing the fractions with $a,b,c$ you get $96 b = x_1 x_2 x_3 x_4, -24 a = x_1 x_2 x_3 x_4, -144 c = x_1 x_2 x_3 x_4$. Since this is what you want your first linear equation is $96 b - 24 a -144 c =1$. Repeat for other combinations of $x_i$ until you have enough independent linear equations to solve for $a,b,c$. E.g. for $x_i^4$ you get $-4b-a-6c=0$.
• Solve for $a,b,c$.

With some combinatorics it should be possible to work out the general rule.

Sketch of a proof for the general rule in granular bastard's answer.

1. The number of sign permutations of the form $\pm x_1 \pm x_2 \pm \dots \pm x_k$ is $2^k$. The number of completely mixed monomials $x_1 x_2 \cdots x_k$ when expanding $(\pm x_1 \pm x_2\pm \dots \pm x_k)^k$ is $k!$. Therefore, the completely mixed monomial occurs $k!\times 2^k=(2k)!!$ times. Due to the construction of the above expression with $(-1)^l=\left(\prod_{i=1}^k s_i \right)$ with $l$ being the number of minuses in each bracket, all completely mixed monomials $x_1 x_2 \cdots x_k$ have a positive sign.

2. In each monomial that is not completely mixed, like $x_1^2 x_3 \cdots x_k$ where $x_2$ is missing, one or more $x_j$ does not contribute. For each missing $x_j$ there is always a pair of terms like $(\pm x_1\dots + x_j \dots\pm x_k)^k$ and $-(\pm x_1 \dots- x_j \dots \pm x_k)^k$. From these, the monomials that do not contain $x_j$ cancel each other out due to the construction of the above expression.

Therefore, only the completely mixed monomials are not cancelled out.

Writing the terms symmetrically, for $k=1,2,3,4$:

$$\begin{align} &(2\times 1)\text{!!}\prod_{i=1}^1 x_i=2x_1=\\ &+\left(+x_1\right){}^1\\ &-\left(-x_1\right){}^1\\ \\ &(2\times 2)\text{!!}\prod_{i=1}^2 x_i=8x_1x_2=\\ &+\left(+x_1+x_2\right)^2+\left(-x_1-x_2\right)^2\\ &-\left(+x_1-x_2\right)^2-\left(-x_2+x_1\right)^2\\ \\ &(2\times 3)\text{!!}\prod _{i=1}^3 x_i=48x_1x_2x_3=\\ &+\left(+x_1+x_2+x_3\right)^3+\left(+x_1-x_2-x_3\right)^3+\left(-x_1+x_2-x_3\right)^3+\left(-x_1-x_2+x_3\right)^3\\ &-\left(-x_1-x_2-x_3\right)^3-\left(-x_1+x_2+x_3\right)^3-\left(+x_1-x_2+x_3\right)^3-\left(+x_1+x_2-x_3\right)^3\\ \\ &(2\times 4)\text{!!}\prod_{i=1}^4 x_i=384x_1x_2x_3x_4=\\ &+\left(+x_1+x_2+x_3+x_4\right)^4+\left(-x_1-x_2-x_3-x_4\right)^4+\left(+x_1+x_2-x_3-x_4\right)^4+\left(+x_1-x_2+x_3-x_4\right)^4 +\left(-x_1+x_2+x_3-x_4\right)^4+\left(+x_1-x_2-x_3+x_4\right)^4+\left(-x_1+x_2-x_3+x_4\right)^4+\left(-x_1-x_2+x_3+x_4\right)^4\\ &-\left(-x_1+x_2+x_3+x_4\right)^4-\left(+x_1-x_2+x_3+x_4\right)^4-\left(+x_1+x_2-x_3+x_4\right)^4-\left(+x_1+x_2+x_3-x_4\right)^4 -\left(+x_1-x_2-x_3-x_4\right)^4-\left(-x_1+x_2-x_3-x_4\right)^4-\left(-x_1-x_2+x_3-x_4\right)^4-\left(-x_1-x_2-x_3+x_4\right)^4 ,\end{align}$$

where !! is the double factorial, we see that the $2^k$ summands are all sign combinations of $x_i$. The signs in front of the brackets are negative/positive if the number of negative signs within the bracket are odd/even. Using a programming language this construction rule could be proved for $1\le k\le28$.

Using the definition described here we can write a conjecture for $k\in \mathcal{N}^+$ $$\prod_{i=1}^k x_i =\frac{1}{(2k)!!} \sum_{(s_1, s_2, \dots, s_k) \in \{-1,1\}^k}\left[\left(\prod_{i=1}^k s_i \right)\left(\sum_{i=1}^k {s_i x_i}\right)^k\right],$$ where $\{-1, 1\}^k := \Big\{(s_1, s_2, \dots, s_k) \mid s_i \in \{-1, 1\} \text{ for all positive integers } i \leq k \Big\}$ are the set of all $2^k$ tuples of length $k$ whose elements are either $−1$ or $1$. So, for example, the set $\{−1,1\}^2$ is defined to be $\{(−1,−1),(−1,1),(1,-1,),(1,1)\}$.

It would be interesting to see a beautiful mathematical proof for $k\in \mathcal{N}^+$.