Why is 0! = 1?
It is defined that way. That's why.
Is there no proof to that?
All proofs have at their basis a set of axioms&definitions, and a proof is simply to show that something else follows from those very same axioms&definitions.
Thus, neither axioms or definitions are themselves things to be proven, although it is quite possible that one may set up OTHER axioms&definitions from which the elements of the first set can be proven.
How would you, for example prove that a+0=a for any number a?
Not that if you have a general recurrence relation described as:
(typically part of the definition of the factorial) you could, if you ASSUME this to be valid for n>=1 insert for n=1:
that is R(1)=R(0).
Now, how are you to go from this to your standard idea of the factorial?
Clearly, by fixing the value R(1)=R(0)=1.
This is therefore a necessary additional definition, since the relation R(n)=n*R(n-1) can have other sequences related to it, for example R(n)=0 for all n.
n! is the number of possible ways to scramble up n objects & there's only one way to scramble up zero objects. It's a bit similar to showing there's only one empty set; if there were another way, what would it look like?
n! might be INTERPRETED as that, if you like.
^ that's how I made it make sense to myself anyway
0! = 1 is the value you get from the Gamma function, many series expansions are more compactly expressed if 0! = 1. The number of permutations of an empty set is 1.
It's simply more convenient for most situations where factorials are used that one defines 0! to be 1.
Separate names with a comma.