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, $0!=1$

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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.

Venkata Karthik Bandaru - 5 years, 7 months ago

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 - 5 years, 7 months ago

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 !

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 - 5 years, 7 months ago

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 - 5 years, 7 months ago

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

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, 6 months ago

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`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$