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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
1 year, 9 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. · 1 year, 9 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? · 1 year, 9 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 ! · 1 year, 9 months ago

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. · 1 year, 8 months ago

Is this proof correct? $n!=n(n-1)!$ $\frac{n!}{n}=(n-1)!$ For $$n=1$$ $\frac{1!}{1}=0!$ So $$0!=1$$ · 1 year, 9 months ago

Definitely no ! The identity $$n! = n(n-1)!$$ for every positive integer $$n$$ is valid because of the assumption $$0! = 1$$. · 1 year, 9 months ago

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. · 1 year, 9 months ago