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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 \(\Gamma(x)\), which is defined by

\[\Gamma(x) = \int_0^\infty \mathrm{t}^{x-1}{e}^{-t}\,\mathrm{d}t\]

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

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

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

Integrating by parts, we have \[\Gamma(n) = (n-1) \int_0^\infty \mathrm{t}^{(n-1)-1}{e}^{-t}\,\mathrm{d}t\]

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

Repeating the above process, \[\Gamma(n)= (n-1)(n-2)(n-3) \dots\dots 1 \times \Gamma(1)\]

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

So, \(\Gamma(n) = (n-1)!\)

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

Note by Ameya Salankar
11 months ago

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As far as I came to know, you are going against the chain of reasoning. \( 0! \) is defined to be \( 1 \), 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. Karthik Venkata · 11 months ago

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@Karthik Venkata 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 \(\Gamma(1) = 1\), one should use Gamma function. This is the beauty of mathematics and we should not harm it. you agree with me? Kartik Sharma · 11 months ago

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@Kartik Sharma I do agree with you.

I did not mean that the Gamma function is derived from the factorial, but meant that the result \( \Gamma(n) = (n-1)! \) is consistent with combinatorial definition of a factorial only because of this assumption that \( 0! = 1 \). In other words, \( 0! = 1 \) 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 ! Karthik Venkata · 11 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 · 9 months, 3 weeks ago

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Is this proof correct? \[n!=n(n-1)!\] \[\frac{n!}{n}=(n-1)!\] For \(n=1\) \[\frac{1!}{1}=0!\] So \(0!=1\) Hjalmar Orellana Soto · 11 months ago

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@Hjalmar Orellana Soto Definitely no ! The identity \( n! = n(n-1)! \) for every positive integer \( n \) is valid because of the assumption \( 0! = 1 \). Karthik Venkata · 11 months ago

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I searched and found that the recurrence formula for the gamma function was \(\Gamma(n+1)=n!\) But then what's wrong with \(\Gamma(n)=(n-1)!\) It doesn't seem to meet any contradiction. Ameya Salankar · 11 months ago

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