Why zero factorial equals one

Well, I can think about 2 more ways why we should define 0!=1

I. n!=1*2*3*4*...*(n-1)*n — this is definition of n!.
n! is multiplication of n factors starting from 1,2,… till n.

For example, 5! is multiplication of 5 factors 1*2*3*4*5 or 5!=1*2*3*4*5.

Similarly, 2! multiplication of 2 factors 1*2 or 2!=1*2.

How much is 1!? 1! is multiplication of 1 factor 1, 1!=1 (you can ask what does multiplication of 1 factor mean? Well, there is “natural” definition. Because n*1=n you can think of multiplication of 1 factor n as simple multiplication n by 1, that is multiplication of 1 factor 1 is m*1=m)

How much is 0!? 0! should be multiplication of 0 factors, so called “empty product”. The value of “empty product” is 1. (Again, it is because n*1=n, so for any product we can just add 1 to the product. Alternatively, it can be derived from “empty sum”) as following:

Since logarithms turn products into sums:

they should map an empty product to an empty sum. If we define the empty sum to be 0, then the empty product should be e⁰= 1.

(Conversely, if we define the empty product to be 1, then the empty sum should be ln ⁡ 1 = 0.).

II. Another way.

  • From the formula n!=1*2*3*4..*(n-1)*n we can define recursive formula as follows:

    For n>2 n!=n*(n-1)! (note, that n-1>1>0)
    For n=2 2!=2
    (and 1!=1 by definion)

    Let try it:

    5!=5*4!
    4!=4*3!
    3!=3*2!
    2!=2 (base case)

    so
    3!=3*2=6
    4!=4*6*24
    5!=5*24=120

    But why we chose as base case n=2? Let put it n=1.
  • For n>1 n!=n*(n-1)! (note, that n-1>0)
    For n=1 1!=1

    So, if from previous derivation we have 2!=2 by definition (of the base case), now we can calculate it:
    2!={2>1}=2!=2*(2–1)!=2*1!={1!=1 by definition of the base case}=2*1*2 that is consistent with the previous definition.

    Can we go one step further? Can we define base case for n=0? Let’s try.
  • For n>0 n!=n*(n-1)! (note, that n-1>-1, that is for n=0, n-1 will be negative, and for n=1, n-1 will be 1)
    For n=0 0!=?

    Well, let’s try to figure out what is 2!

    2!={2>0}=2*(2–1)!=2*1!
    1!={1>0}=1*(1–1)!=1*0!=0!


    In order that this definition will be consistent with previous ones above, we should define 0!=1. That is we should define:

    For n>0 n!=n*(n-1)! (note, that n-1>-1, we can’t put n=0 to this formula, only n>0)
    For n=0 0!=1

    In this case
    1!=0!=1 as expected.

Originally published in my blog https://www.toalexsmail.com/2012/04/why-zero-factorial-equals-one-english.html

Senior Software Engineer at Pursway

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