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?


I think that the definition $0!=1$ is the one which makes most of the formulas work nicely.
For example $(n+1)!=n! (n+1)$ also works for $n=0$, as long as $0!=1$.
But most importantly, $\binom{n}{k}$ also works in the case $k=0$. Note that $\binom{n}{0}$ has to be $1$, if you want to have a nice formula for $(a+b)^n$. So for convenienceence, we define $\binom{n}{0}$ to be $1$, and if you want the binomial coefficient to be given by the standard formula, then $0!$ has to be 1.
