Conjecture: If $A$ is a nonempty, finite set with size $|A|=n$, and $$P_A(x)=\prod_{a\in A}(x-a),$$ then $P_A(x)$ can be expanded as $$P_A(x)=\sum_{k=0}^{n}(-1)^{n-k}x^k\sum_{{U\subseteq A}\atop{|U|=n-k}}\prod_{u\in U}u.\tag{1}$$
I have conjectured this based on algebraic evidence. That is, I expanded out the cases $n=1,...,4$ both manually and through $(1)$, and in each case the conjecture held. The problem is, I'm having difficulty proving this result. I'm fairly certain that a proof would involve induction, but I'm not very good at that, and have so far failed.
I was initially interested in this formula because I recognized that $(1)$ would imply that $$k\ne0,n\iff \sum_{{U\subseteq S_n}\atop{|U|=k}}\prod_{u\in U}u=0\tag{2}$$ for $$S_n=\left\{\exp\frac{2\pi ik}{n}:k=0,1,...,n-1\right\}.$$ It may seem un-intuitive at first, but $(2)$ is true because $$P_{S_n}(x)=x^n-1,$$ (as $S_n$ is the set of roots of $x^n-1$) so each term of the expansion must vanish except for the cases $k=0,n$.
After seeing this, I was naturally curious about a proof of $(1)$. Could I have some help or hints? Thanks.
Yes, you have discovered the elementary symmetric polynomials and their relationship with the coefficients of a given polynomial (see Section 3, Properties, at the link above).