There are lots of perfectly good GQs posted by still-active posters which never got a response. I can only hope that one day some benevolent Doper will take an interest in one of the 0-reply questions I’ve posted and provide the information I’ve desperately been searching for for so long. 
Yeah, but how did you stumble on this one?
The Gamma function, on whatever particular calculus-y construction, doesn’t really add anything to the argument. Yes, defining n! as Gamma(n + 1) yields 0! = 1, but the only reason anyone uses this definition of n! is because it, among other things, satisfies the property that 1! = 1 and n! = (n - 1)! * n. But those properties yield 0! = 1 rather directly and simply just in themselves. Sure, this is reflected in the Gamma function, but that’s just because the Gamma function, by design, satisfies those properties as well.
There is no argument better than “1! = 0! * 1 [in keeping with the defining n! = (n - 1)! * n pattern]; therefore, 0! = 1!/1 = 1”. That’s all there is to it, and all there need be.
I could swear I edited the wording on this line, but somehow it got clobbered by a later edit?
Missing words reinstated in bold.
Also, as far as I’m concerned, we should just go ahead and feel comfortable writing x! even for non-integer x, or write \Pi(x) or what have you, all the time, rather than ever bothering with \Gamma(x - 1). The shift by one in the definition of Gamma that most people use has no great motivation and just causes needless obscuration.
As long as we’re going to talk about mysteries of posting, why did jeturgeon post six times in 2000, where three of the posts are duplicates of each other, and then not post again till this year?
I’ve always found the permutation motivation compelling. The number of ways to order no objects is 1, namely, having no objects.
Yes, but I’ve noticed that many find “The number of ways to order no objects is 1” as confusing a concept as “0! = 1” or “the empty product is 1” or what have you.
At any rate, that again can be seen as just a consequence of the fact that the number of ways to order n objects, for finite positive n, is equal to the number of ways to order (n - 1) objects times n [as in, first ordering (n - 1) of the objects and then picking one of n positions in which to place the remaining object]. So, again, we are back to 0! = 1!/1 = 1.
No; you may perfectly well go ahead and take (-1)! to be 0!/0 = 1/0 = unsigned infinity. After all, the meaning of -1 x -2 -x -3 x … x 1 is far from immediately clear, and no one ever guaranteed it would come out nice. Indeed, the factorial function blows up in the same way at all negative integers, precisely because of that division by 0.
[And then the argument “0! = 0 x (-1)! = 0” fails, because while ordinarily 0 times anything is 0, once we acknowledge (-1)! as “blowing up”, we’re looking at the indeterminate form 0 x unsigned infinity, which needn’t be 0 and in this case is describing 1 instead]
I suppose I meant \Gamma(x + 1). But, that’s the point; having to keep track of this shift makes things needlessly difficult.
Anyway, speaking of the Gamma function, for those who’ve never seen a good account of how the factorial function extends naturally to non-integers, let me present one:
For integers a and b, let [a, b] denote b!/a! = b x (b - 1) x (b - 2) x … x (a + 1).
Note that this has the following fundamental properties:
[a, a] = 1
[a, c] = [a, b] x [b, c]
(From these, we also have [a, b] = 1/[b, a])
Also,
[a - 1, a] = a
And finally, the one perhaps trickier property to keep in mind:
If the difference between a and b is held fixed while a and b grow large, then [a, b] is asymptotically equal to a^(b - a). [This is because a^(b - a) <= [a, b] <= b^(b - a), but the bounds a^(b - a) and b^(b - a) are asymptotically equal to each other. (This, in turn, is because, the bounds’ ratio is (1 + (b - a)/a)^(b - a), which approaches (1 + (b - a) x 0)^(b - a) = 1 as a grows large]
Using all these properties together, we have that [a, b] = [a, a + n] x [a + n, b + n] x [b + n, b] = [a + n, b + n] x [a, a + n] / [b, b + n] = [a + n, b + n] x ((a + n) x (a + n - 1) x … x (a + 1))/((b + n) x (b + n - 1) x … x (b + 1)), no matter what integer n is. And, in the limit as n goes to infinity, [a + n, b + n] becomes asymptotically equal to a^(b - a). Thus, we have that [a, b] = the limit as the integer n goes to infinity of (a + n)^(b - a) x ((a + n) x (a + n - 1) x … x (a + 1))/((b + n) x (b + n - 1) x … x (b + 1)).
But this last definition can be interpreted even if a and b aren’t integers. And that gives the canonical extension of factorial to non-integer inputs (aka, the gamma function).
Typo corrected in bold.
And, although I didn’t say so explicitly, I should: I spoke above in terms of the general product [a, b] = b x (b - 1) x … x (a + 1), but the way to recover ordinary factorial from this is of course to interpret n! as [0, n].
This is really the best explanation. All the rest of the explanations just say that you want 0! to equal 1 in order to make a formula work that is really only meaningful for n>1, and that is not a very compelling reason.
Why say the formula is only meaningful for n > 1? The formula is meaningful for all n, including 0, negative integers, and even non-integers. It’s just that the factorial function hits unsigned infinity on negative integers, and is more complicated to understand on non-integers. But the defining pattern captured by the formula is valid everywhere.
[If one fundamentally thinks of n! as “The number of ways to arrange n objects”, then of course the answer to questions such as the OP’s must be “The number of ways to arrange 0 objects is 1”, with some further explanation necessary for those who have difficulty thinking about “The number of ways to arrange 0 objects”. But I think most people asking questions such as the OP’s don’t fundamentally think of n! as “The number of ways to arrange n objects” at all, but rather, as simply “The function which keeps multiplying by the next number as you go on”. And this is perfectly fine; after all, it’s this latter pattern which we use to fruitfully generalize the factorial function quite far, while the definition in terms of numbers of orderings is limited to making sense only for cardinal numbers]
Why wouldn’t that be compelling? After all, math, and especially the way we define functions, is for our convenience.
The function Factorial(n) = n! the way it is usually defined shows up in a lot of useful situations. The function Schmactorial(n) = n! for n >= 1 and something else for n = 0 doesn’t show up in many places at all. So if we defined our series in terms of Schmactorial() instead of Factorial() we’d have to pull out a lot of special cases, and write a lot of equations less compactly, &c, and it would get tedious.
That’s the reason I cite for 1 not being a prime. If 1 was a prime, then a lot of useful theorems that now start with “let p be a prime” would then say “let p be a prime that is not one”, and that would suck.
No, the Gamma Function comparison doesn’t rely on "making a formula work that is only meaningful for n > 1.
Oh, I don’t know about that. Good luck trying to convince anyone that n! ought mean, say, “The integral from 0 to infinity of t^n/e^t dt” without resort to exhibiting the recurrence relation formula [though I agree that formula is not only meaningful for n > 1].
Damn, math zombie, you scary!