The much simpler answer is that this is the special case of the empty product being 1 (or more precisely, the neutral element of the monoid on which your multip

The much simpler answer is that this is the special case of the empty product being 1 (or more precisely, the neutral element of the monoid on which your multiplication operates). The empty product is 1 so that $\prod_{x \in S \setminus T}x\prod_{x \in T}x = \prod_{x \in S}x$ holds for all $T \subseteq S$. It's the exact same reason why the empty sum is zero (zero being the neutral element of a monoid using additive notation). n! is commonly defined as the product of all positive integers less than or equal to n. For n = 0, this is the product of the empty set.