How can I reconcile the 1 on the LHS of $\prod\limits_{i \in I}(1+x_i)=\sum\limits_{J \subseteq I}\prod _{j \in J} x_j$? Byron Schmuland used this equality in his answer: $\prod _{i \in I}(1+x_i)=\sum_{J \subseteq I}\prod _{j \in J} x_j$.
I can see all the combinations being replicated on both sides of the equality, however I cannot reconcile the '1' that results from the $\prod (1+x_i)$ on left side.
 A: Proving a more general formula may simplify the proof: consider the  product
$$\prod_{i\in I}(e_i+x_i)$$
By the distributive law, you have to take the product of  $x_i$s in a number of factors (the indices thereof form a subset $J$ of the  indexing set $I$) and $e_i$s in the remaining factors (their indices form the  subset $I\smallsetminus J$), in all possible ways, i.e. $J$ can be any subset of $I$, and add these products. Thus we obtain:
$$\prod_{i\in I}(e_i+x_i)=\sum_{J\subset I}\biggl(\prod_{i\in I\smallsetminus J}e_i\prod_{j\in J}x_j\biggr).$$
Now set $e_i=1$ for all $i$, and you get your formula.
A: It is convenient to see this in a particular case. Suppose that $I=\{1,2,3\}$. Then$$\prod_{i\in I}(1+x_i)=(1+x_1)(1+x_2)(1+x_3)(1+x_4).$$On the other hand, the subsets of $I$ are $\emptyset$, $\{1\}$, $\{2\}$, $\{3\}$, $\{1,2\}$, $\{1,3\}$, $\{2,3\}$, and $\{1,2,3\}$. So\begin{align*}\sum_{J\subset I}\prod_{j\in J}x_j&=\prod_{j\in\emptyset}x_j+\prod_{j\in\{1\}}x_j+\prod_{j\in\{2\}}x_j+\prod_{j\in\{3\}}x_j+\prod_{j\in\{1,2\}}x_j+\prod_{j\in\{1,3\}}x_j+\prod_{j\in\{2,3\}}x_j+\prod_{j\in\{1,2,3\}}x_j\\&=1+x_1+x_2+x_3+x_1x_2+x_1x_3+x_2x_3+x_1x_2x_3\text,\end{align*}which is equal to the first product.
It is easy to provide a formal proof of this equality in the general case by induction.
