$\prod_{n=0}^\infty(1+x_m)=\sum_{S\in\mathcal{P}_f(\mathbb{W})} \prod_{m\in S}{x_m}$ I am self studying products and had some success exploiting (slight variants) of the infinite analog of 
$$ \prod_{n=0}^m (1+x_n)=\sum_{S\subset [m]}\prod_{i\in S} x_i $$ 
$[m] =\{0,1,2,\dots, m\}$  to make a claim about the zeta function as well in some other places. And I just want to suss out the details here: What happens as we take $m\to \infty$? 
Question: Is it the case that:
$$\prod_{n=0}^\infty(1+x_n)=\sum_{S\in\mathcal{P}_f(\mathbb{W})} \prod_{j\in S}{x_j}$$  
Where $\mathcal P_f(\mathbb W)$ is the set of finite subsets of the non-negative natural numbers. It feels like this has just gotta be the case. We have been warned to avoid the "obvious trap": which is to say "Oh yeah, that's obvious."
I can't quite see through the details of showing this formally. Thanks for any help. A good push in the right direction would be welcome. I wouldn't mind if someone wanted to simply prove it outright. 
 A: I assume that $\mathbb{W}$ denote the set $\mathbb{Z}_{\geq 0}$ of nonnegative integers.  I shall prove that, for real numbers $x_0,x_1,x_2,\ldots$, we have
$$\prod_{m\in\mathbb{W}}\,\left(1+x_m\right)=\sum_{S\in \mathcal{P}_f(\mathbb{W})}\,\prod_{s\in S}\,x_s\,,\tag{*}$$
provided that $\prod\limits_{m\in\mathbb{W}}\,\left(1+x_m\right)$ converges unconditionally to a nonzero limit.  

 By unconditional convergence to a nonzero limit, I mean that, for any renumeration $\{t_0,t_1,t_2,\ldots\}$ of $\mathbb{W}$, we have that $$\lim_{N\to\infty}\,\prod_{m=0}^N\,\left(1+x_{t_m}\right)$$ converges to a nonzero limit.  I forgot about unconditional convergence in my comment, and it should be clear why we need this extra condition.  However, if $x_0,x_1,x_2,\ldots$ can be complex numbers, then a good condition is that the product $\prod\limits_{m\in\mathbb{W}}\,\big(1+|x_m|\big)$ converges (i.e. $\prod\limits_{m\in\mathbb{W}}\,(1+x_m)$ converges absolutely).  See also this thread.  If $x_0,x_1,x_2,\ldots$ are real numbers, then $\prod\limits_{m\in\mathbb{W}}\,\left(1+x_m\right)$ converges unconditionally to a nonzero limit if and only if $x_m\neq -1$ for all $m\in\mathbb{W}$ and $\prod\limits_{m\in\mathbb{W}}\,\left(1+x_m\right)$ converges absolutely.

Suppose that  $\prod\limits_{m\in\mathbb{W}}\,\left(1+x_m\right)$ converges unconditionally and $$\lim_{N\to\infty}\,\prod_{m=0}^N\,\left(1+x_m\right)=L\,,$$
where $L\neq 0$.  First of all, (*) is clearly true if $x_m\geq 0$ for every $m\in\mathbb{W}$.  Our strategy is to consider $m\in\mathbb{W}$ such that $x_m\geq 0$ and $m\in\mathbb{W}$ such that $x_m<0$. 
As the product converges unconditionally, we have
$$\prod_{m\in\mathbb{W}^+}\,\left(1+x_m\right)=L^+\text{ and }\prod_{m\in \mathbb{W}^-}\,\left(1+x_m\right)=L^-\,,$$
for some $L^+>0$ and $L^-\in\mathbb{R}\setminus\{0\}$ (with $L=L^+L^-$).  Here, $$\mathbb{W}^+:=\left\{m\in\mathbb{W}\,\big|\,x_m\geq 0\right\}\text{ and }\mathbb{W}^-:=\left\{m\in\mathbb{W}\,\big|\,x_m<0\right\}\,.$$
Now, you can handle the products $\prod\limits_{m\in\mathbb{W}^+}\,\left(1+x_m\right)$ and $\prod\limits_{m\in\mathbb{W}^-}\,\left(1+x_m\right)$ separately.  For $\prod\limits_{m\in\mathbb{W}^+}\,\left(1+x_m\right)$, we clearly have a version of (*):
$$\prod_{m\in\mathbb{W}^+}\,\left(1+x_m\right)=\sum_{S\in\mathcal{P}_f(\mathbb{W}^+)}\,\prod_{s\in S}\,x_s\,.$$  
For $\prod\limits_{m\in\mathbb{W}^-}\,\left(1+x_m\right)$, we may without loss of generality assume that $x_m>-1$ for all $m\in\mathbb{W}^-$ (otherwise, note that there are finitely many $m$ such that $x_m<-1$, and we can remove them).  That is, $L^->0$, and so
$$\frac{1}{L^-}=\prod_{m\in\mathbb{W}^-}\,\left(\frac{1}{1+x_m}\right)>\prod_{m\in\mathbb{W}^-}\,\left(1-x_m\right)>0\,.$$
Thus, $\prod\limits_{m\in\mathbb{W}^-}\,\left(1-x_m\right)$ converges, and we get a version of (*) for $\prod\limits_{m\in\mathbb{W}^-}\,\left(1-x_m\right)$:
$$\prod_{m\in\mathbb{W}^-}\,\left(1-x_m\right)=\sum_{S\in\mathcal{P}_f(\mathbb{W}^-)}\,\prod_{s\in S}\,(-x_s)\,.$$
This shows that $\sum\limits_{S\in\mathcal{P}_f(\mathbb{W}^-)}\,\prod\limits_{s\in S}\,x_s$ converges absolutely, whence
$$\prod_{m\in\mathbb{W}^-}\,\left(1+x_m\right)=\sum\limits_{S\in\mathcal{P}_f(\mathbb{W}^-)}\,\prod\limits_{s\in S}\,x_s\,.$$
