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© Copyright 1999, Jim Loy
n! (n factorial) is defined to be the product of all of the positive integers from one, up to n. In other "words," n!=1x2x3x...xn:
1!=1
2!=2
3!=6
4!=24
5!=120
6!=720
7!=5040
8!=40,320
9!=362,880
...
Factorials get large, fairly fast. They are very important in probability and calculus.
We sometimes see that 0!=1. Why is that? It would seem that we cannot do the multiplications from one "up to" zero. And, if we go backwards (1x0), we get zero.
Experimenting with factorials, we come up with n!=n(n-1)!. For example 17!=17x(16!):
16!=1x2x...x16
17!=(1x2x...x16)x17
That equation (n!=n(n-1)!) just dictated to us where to put the parentheses. By making n=1, we can find 0!:
1!=1(0!)
0!=1
And, it turns out that 0!=1 works very well in many situations (in probability, for example).
Addendum:
The above proof that 0!=1 is based upon n!=n(n-1)!, which is in turn based upon the definition of factorial. So, it would seem to be a valid proof. But, 0! cannot be defined directly from the definition of factorial. So, mathematicians like to define 0! as 1, without proving it. So, the proof just amounts to a demonstration that defining 0! as 1 is consistent with the definition of factorial.
I received email saying that my proof that 0!=1 is invalid because 0! is a constant, and you cannot solve for a constant. Wrong. I can find 4! (in a number of ways), and 4! is a constant. I can solve for the square root of 7, probably using my calculator, and it is a constant, too. There are numerous ways of finding pi, and it too is a constant.
n! is also the number of permutations (ways of arranging) exactly n things. It makes sense to say that there is one way to arrange zero things. So again, 0!=1.