How do you prove that 0!=1?

[Dick]

If I understand correctly, you define n!=Γ(n+1) for all integer and I define n!=n(n-1)...1. In the scope of my definition, 0! is NOT defined.

Depending on the OP's definition of n!, either the Γ function argument is (obviously) a satisfactory answer to his question, or there is no absolute answer because it is a matter of convention. In this later case, agreed, the Γ function argument would be one among many other justifications of the convenient convention 0!=1.

Considering "The attempt at a solution" of the OP, it seemed obvious to me that he was working in the set of integers, and that he was looking for an (impossible) arithmetic proof. I agree that your analytic definition is on top of the pure arithmeric one, but it does rely on a much heavier set of axioms.

Apologies if I'm starting to sound like a troll :)
Cheers

Everybody is right here. So this is beating an old dead horse. Defining n!=n(n-1)...1 clearly makes no sense if n=0. Making 0! an arbitrary definition. But then the notation n!=n(n-1)...1 is also vague. Best to define it inductively by (n+1)!=n!*(n+1). But where to start the induction? If you start it by defining 1!=1 that works. But you could also start it by defining 0!=1 and get the same function. Clearly there's a problem with defining (-1)! equal to some number and trying to proceed from there. Starting by defining 0!=1 and going inductively from there agrees with the gamma function definition and gives you the usual definition of n!. So why not 0!=1?