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...
Indeed. I'm a new joiner. This post was on top of some category.
But, not matter how old the thread is, it is very frustrating that it did not meet its purpose : some high school student asked why 0! = 1 ; most replies were tentative demonstrations, hence gibberish. The answer the OP needed had...
You've been told to accept it as true because it is a convenient convention. Just like any other word, n! means nothing unless we agree on a meaning for it. Well, n! means n*((n-1)!) if n >0 and 0! means 1. People could have agreed that n! means n*((n-1)!) if n >1, that 1! means 1, and that 0...
lol, I get it now. Not "prove that 0 is different from 1", but "prove that factorial of 0 is 1".
Just refer to the definition of factorial, then, 0! = 1 does not contain ANY information whatsoever.
Agreed. There are some sets containing X such that X*a=a (X==1) and X+a=a (X==0) for all a. Depending on the definition of '*' and '+', I suppose there could be more than one such set.
So, I don't know what the OP means, really. I mean, if we're talking Peano's integers, then 1!=0 because 1 is...