Can someone pleeeease explain this for me. Throughout all of my math classes, mostly Algebra, I was always told that zero factorial (0!) is equal to 1. Now I understand how 3! is the same as 6, and 5! is the same as 120, and so forth, but I cannot figure out the 0! mystery. Any help would be appreciated.
First, take the definition of the factorial function:
n! = n × (n − 1) × (n − 2) × ⋯ × 1
If you agree with this, then you should agree that 1! = 1; we simply substitute 1 for n in the above definition.
The definition says what the factorial of n is, so we should be able to find out what it is when applied to (n − 1):
(n − 1)! = (n − 1) × ((n − 1) − 1) × ((n − 1) − 2) × ⋯ × 1
So simplifying we get
(n − 1)! = (n − 1) × (n − 2) × ⋯ × 1
Since the right-hand side of this equation already appears in our definition of n!, we can substitute the left-hand side there:
n! = n × (n − 1)!
So now let’s evaluate this slightly modified definition for n = 1:
1! = 1 × (1 − 1)!
Subtracting, we get
1! = 1 × 0!
And since we already established that 1! = 1, we can make a substitution in the left-hand side of this equation:
1 = 1 × 0!
And now we can divide both sides by 1 to cancel out the 1 on the right-hand side:
1 ÷ 1 = (1 × 0!) ÷ 1
So simplifying,
1 = 0!
It also turns out that many infinite series and sequences that include the term n! can be extrapolated down to ** n = 0 ** if n! = 1.
But the biggest reason is that the factorial function follows the Gamma function. Gamma(n) is equal to (n - 1)!, and Gamma (1) = 1
If n! = 1 you’ve got a very interesting and very simple (but not very useful) definition of the factorial function. 
Here’s what I find paradoxical. As the above is, as you say, the definition of the factorial function, let us put in 0 for n to find 0!.
0! = 0 x (0 - 1) x (0 - 2) x — x 1
This simplifies to 0! = 0
What am I missing here?
Edit: Have I been whooshed?
One thing to keep in mind is that factorials are about multiplication, and with multiplication, the “starting point” or “do-nothing default” is 1, not 0.
There are many useful rules and relationships that only work consistently if we define 0! to be 1, not 0. For example, n! / n = (n-1)!
Well, yeah, psychonaut’s definition doesn’t apply when n = 0. It basically says to multiply together all the whole numbers from n down to 1—but “n down to 1” doesn’t make sense when you start with 0.
Psychonaut answered it, but to give a bit more perspective: it’s similar to the question “what happens when you multiply no numbers together?”
This sounds like an absurd question, but it’s actually quite reasonable. 2[sup]3[/sup]=222=8, 2[sup]2[/sup]=2*2=4, 2[sup]1[/sup]=2, 2[sup]0[/sup]=?.
As with 0!, the answer here is 1. It’s the only way to make sense of x[sup]0[/sup], for one thing, and also plays nicely with lots of other math. You can convince yourself of it with a similar argument to the factorial one:
x[sup]n[/sup] = x * x[sup]n-1[/sup]
x[sup]1[/sup] = x * x[sup]0[/sup]
x[sup]0[/sup] = x[sup]1[/sup] / x = x / x = 1 (assuming x != 0)
I think that when he says “… x 1”, that implies that it only holds for n>=1 (e.g., 1! isn’t equal to 1 X 0 X -1 X-2…), which would simplify to 1! = 1 x 1.
Edit…I bet the OP is thrilled that’s he’s finally getting a response, even though it took a while
Sure it does. The product of zero numbers is one—it’s known as the empty product.
Yeah. He’s still active here so I figured he might appreciate the information despite the delay. 
YEah, but that’s not in your definition; you had:
“n! = n × (n − 1) × (n − 2) × ⋯ × 1”
Which doesn’t define n! for n<1. So, you should probably have:
n! = n × (n − 1) × (n − 2) × ⋯ × 1, 0! = 1
Edit: missed your response above my post
So then, what’s (-1)! ?
Heck, for that matter, what’s 0.5! ?
-1! is undefined. 0.5! is about 0.8862, and comes from the definition of the Gamma function.
ETA: 0.8862 is sqrt(pi)/2
You’re perhaps interpreting the notation too literally; by your reading it doesn’t properly define n! for n = 1 or n = 2 either, since in the first case you end up multiplying by −1 and in both cases by 0. I could have instead written something like
n! = ∏ {i | i ∈ ℤ[sup]+[/sup] ∧ i ≤ n}
but I couldn’t be sure the OP would understand that notation. It looks like his experience was with high school-level algebra, so I used high school algebra-level notation.
In practical terms, n! is the number of ways of arranging n items in a row.
You can arrange 3 items in a row six different ways. Two items in two ways. One item in only one way.
And zero items? Only one way!
Psychonaut, how did this zombie happen?
This is certainly correct, but it can be put into a broader context. Here is the nerd’s eye view: Generally, if S and T are two sets of number, assumed disjoint (no elements in common), then the product of all the numbers in S times the product of all the numbers in T is the same as the product of all the numbers is S \cup T (the union of S and T). Let S be empty and you conclude the product over the empty set has to be 1. Incidentally, if S and T have common elements, the product of all the numbers in S \cup T times the product of all the numbers in S \cap T (the intersection) is the product of all the numbers in S times the product of all the numbers in T. If this formula is to remain valid when S and T are disjoint, then again the empty product is 1. The fact that the Gamma function agrees with this is nice, but the combinatorial explanation is decisive. By the way, combinatorially, 0^0 is also 1 and in no way undefined.
The same argument with sum replacing product should convince you that the empty sum is 0.
He was showing his work. He works really slow.
Surely this is some kind of record! An OP with no responses, revived TWELVE years later?
I was in a hurry. It’s obviously 0! or n = 0.
But you knew that.