Proof of 0! = 1

0! = 1 can be proved using many methods. While some use patterns or logic, giving reasons for mathematical consistency & this & that, I shall use calculus here.

We have the gamma function denoted by Γ(x)\Gamma(x), which is defined by

Γ(x)=0tx1etdt\Gamma(x) = \int_0^\infty \mathrm{t}^{x-1}{e}^{-t}\,\mathrm{d}t

This function arose while solving an interpolation problem. The problem was to find a monotonic function defined over (1,)(1, \infty) which took the value n!n! at nn. It can be solved by evaluating the above improper integral which converges for x>0x>0.

The gamma function is defined for the whole of real line provided we take Γ(x)= \Gamma(x)=\infty for x=0,1,2,x=0,-1,-2,\dots.

We are interested in the case when x=nx=n that is, a positive integer.

Integrating by parts, we have Γ(n)=(n1)0t(n1)1etdt\Gamma(n) = (n-1) \int_0^\infty \mathrm{t}^{(n-1)-1}{e}^{-t}\,\mathrm{d}t

Notice that the integral is nothing but Γ(n1)\Gamma(n-1). Thus we have, Γ(n)=(n1)×Γ(n1)\Gamma(n)= (n-1) \times \Gamma(n-1).

Repeating the above process, Γ(n)=(n1)(n2)(n3)1×Γ(1)\Gamma(n)= (n-1)(n-2)(n-3) \dots\dots 1 \times \Gamma(1)

But, Γ(1)=0tx1etdt\Gamma(1) = \int_0^\infty \mathrm{t}^{x-1}{e}^{-t}\,\mathrm{d}t =et10=\left.\frac{e^{-t}}{-1}\right|_0^{\infty} =(01)=1=-(0-1) = 1

So, Γ(n)=(n1)!\Gamma(n) = (n-1)!

This must imply that Γ(1)=0!\Gamma(1) = 0! and we have already proved that Γ(1)=1\Gamma(1)=1. Thus, 0!=10!=1

Note by Ameya Salankar
5 years, 8 months ago

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As far as I came to know, you are going against the chain of reasoning. 0! 0! is defined to be 1 1 , and hence the results follow. Just like falling into the trap of circular definitions, one shouldn't try to prove a fundamental definition by using stuff derived/dependent on itself.

Do check Wolfram-MathWorld and Wikipedia links.

Venkata Karthik Bandaru - 5 years, 8 months ago

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I believe that you are right and you are wrong. While you're right when you say that "0! is defined to be 1", I disagree with your connection of Gamma function as "derived from factorial". It is "independent" but yeah, it would be correct to say that to prove 0! = 1, one should not use Gamma function , while when one says that one need to prove Γ(1)=1\Gamma(1) = 1, one should use Gamma function. This is the beauty of mathematics and we should not harm it. you agree with me?

Kartik Sharma - 5 years, 8 months ago

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I do agree with you.

I did not mean that the Gamma function is derived from the factorial, but meant that the result Γ(n)=(n1)! \Gamma(n) = (n-1)! is consistent with combinatorial definition of a factorial only because of this assumption that 0!=1 0! = 1 . In other words, 0!=1 0! = 1 was a convention taken to maintain this consistency.

A same function having two different definitions among different branches of mathematics always leads to such confusion about the order of reasoning !

Venkata Karthik Bandaru - 5 years, 8 months ago

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I searched and found that the recurrence formula for the gamma function was Γ(n+1)=n!\Gamma(n+1)=n! But then what's wrong with Γ(n)=(n1)!\Gamma(n)=(n-1)! It doesn't seem to meet any contradiction.

Ameya Salankar - 5 years, 8 months ago

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Is this proof correct? n!=n(n1)!n!=n(n-1)! n!n=(n1)!\frac{n!}{n}=(n-1)! For n=1n=1 1!1=0!\frac{1!}{1}=0! So 0!=10!=1

Hjalmar Orellana Soto - 5 years, 8 months ago

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Definitely no ! The identity n!=n(n1)! n! = n(n-1)! for every positive integer n n is valid because of the assumption 0!=1 0! = 1 .

Venkata Karthik Bandaru - 5 years, 8 months ago

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0!=1 is a mathematical definition, in order to maintain consistency in the mathematical structure. It cannot be proved and any attempt to prove it is somewhat cyclic in nature.

Kuldeep Guha Mazumder - 5 years, 7 months ago

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