Why 0!=1

From the factorial recurrence relationship of n!=n×(n-1)! and 1!=1:

1=1!=1×0! imples 0!=1 (since identically 1×x =x for all numbers x).

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Factorials

Hopefully the reader is familiar with the counting properties of the factorial that follows the recurrence relation of

  • n!=n (n-1)!
  • 1!=1.

From the above it is not hard to show that:

  • 2!=2(1!)=2(1)=2
  • 3!=3(2!)=3(2)=6
  • 4!=4(3!)=24
  • 5!=5(4!)=120
  • 6!=6(5!)=720

Perhaps a bit complexing is the value of the factorial at 0, this is best viewed as

  • 1!=1(0!)
  • 1=1(0!).

It follows naturally that 0!=1. {author note, ref linked.com group posting via India.}

As gleanned from the unit shifted gamma funciton, .i.e.\Gamma(x+1)=x!, the factorial is never zero, but has poles at the negative integers {-1,-2,-3, … }.

It can be shown that the reciporical of the factorial function is entire. In addition, it has zeros at the negative integers. It is well known that

\Gamma(s) = \frac{1}{s}e^{-\gamma s}\prod_n \exp(s/n)\frac{s}{s+n}

The reciporical factioral (1/s!)! can be shown to satisfy the recurrence of

\frac{1}{(s-1)!} = s \frac{1}{s!}.

  • \frac{1}{(-1)!} = 0 \frac{1}{0!} =0
  • \frac{1}{(-2)!} = -1 \frac{1}{(-1)!} =0
  • \frac{1}{(-3)!} = -2 \frac{1}{(-2)!} =0

ad infinitium. Therefore,

\frac{1}{s!} = e^{\gamma s} \prod_n e^{\frac{-s}{n}}\frac{s+n}{n},

where the product is taken over the naturals.

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