Why $0!$ is equal to $1$? Many counting formulas involving factorials can make sense for the case $n= 0$ if we define $0!=1 $; e.g., Catalan number and the number of trees with a given number of vetrices. Now here is my question:

If $A$ is an associative and commutative ring, then we can define an
  unary operation on the set of all the finite subsets of our ring,
  denoted by $+ \left(A\right)  $ and $\times \left(A\right)$. While it
  is intuitive to define $+ \left( \emptyset \right) =0$, why should
  the product of zero number of elements be $1$? Does the fact that $0! =1$ have anything to do with 1 being the multiplication unity of integers?


In general, we want the "associative" law to hold in the form:
$$
\times(A \cup B) = (\times A) (\times B)
$$
whenever $A$ and $B$ are disjoint.  What would this mean when $B = \emptyset$ ??
