February 4

[Xeroc]

Originally Posted by Derek

^ I liked the proof I heard for 0!.

If 3! = 3*2*1 = 6, and 2! = 2*1 = 2, then 2! can also be expressed as 3!/3. Thus any n! = (n+1)!/(n+1), which is a rather redundant thing to say since really n! is just the product of all the numbers from 1 to n, right?

But what about 0!?! (Will Zeke kill me for that?) Multiplying the numbers from 1 to 0 implies the answer would either be 0, or undefined if 1 must increment. But using the other formula, we see that 0! = (0+1)!/(0+1) = 1!/1 = 1. And sure enough, 0! = 1. But I've always found that proof kind of silly.

There's another one I've heard like this:

Since N! = (N)(N-1)(N-2)...(3)(2)(1)
And (N-1)! = (N-1)(N-2)...(3)(2)(1)
Then N! = (N)(N-1)!
If N = 1
Then 1! = (1)(0)! = 1
So, 0! = 1

! = ()()!