Why are 0! = 1? Shouldn't 0! = 0

No, and here is the reason:

First of all we cant find 0! from the definition which is n! = n(n - 1)(n - 2)(n - 3) ..........*3*2*1
because when you try it with 0!, you can never get to 1 if you keep multiplying by 1 less: 0 * (-1) * (-2)...... (doesnt get to 1). Also some things are unclear, when the numbers are less than 1 do you decreese by 1 or increase? If you increase, you get 0! = 0 * 1 = 0, (-1)! = (-1)*0*1 = 0 etc. Since the method is not obvious, we have to take a different approach.

One of the properties of factorials is this:

n! = n * (n - 1)!

this makes sense since it can be derived from its definition:

n!
= n(n - 1)(n - 2)(n - 3) ..........*3*2*1
= n * [(n - 1)(n - 2)(n - 3) ..........*3*2*1]
= n * (n - 1)!

now rearranging this identity we get:

n! = n * (n - 1)!
(n - 1)! = n! / n

With this identity, we can find 0! by setting n - 1 = 0, so n = 1:

(n - 1)! = n! / n
0! = 1! /1 --------> which we can calculate
0! = 1

also, if we set n - 1 = -1, n = 0:

(-1)! = 0! / 0
(-1)! = infinity (so it is undefined there)

when n - 1 = -2, n = -1:

(-2)! = (-1)! / -1
(-2)! = infinity / -1 (so this is also undefined)


This method to extend factorials to zero and negative numbers ensure that they fulfil the fundamental properties of factorials. However, for negative numbers they are undefined. If you are interested, you can also take a look at the Gamma function which extends factorials to real numbers (not just integers).

http://en.wikipedia.org/wiki/Gamma_funct…

@Robert_W: x^0 isnt something you "just learn". It can be derived logically from the definition of exponents so that all the identities are consistent:

firstly: if n and m are positive integers and let m > n:

x^m / x^n = x*x*x*x*x*x....x (m amount) / x*x*x..x(n amount)
n amount of x cancel out leaving (m - n) amount of x in the top which can be written as x^(m - n). Therefore: