Some people says that this can be demonstrated $0!=1$, but other say that this is a definition. Which one is correct?

Let's given $n\in\mathbb{N}$:

$$(n+1)!=(n+1)\cdot n!$$ $$(0+1)!=(0+1)\cdot 0!$$ $$1!=1\cdot 0!$$ $$1=0!$$

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    $\begingroup$ It depends. What is the definition of factorial? $\endgroup$ – Hagen von Eitzen Oct 29 '14 at 21:00
  • $\begingroup$ It's asumed that the definition is $(n+1)!=n!\cdot (n+1)$. $\endgroup$ – hkviktor Oct 29 '14 at 21:05
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    $\begingroup$ $(n+1)!=n!\cdot (n+1)$ is not enough, as you have to start somewhere. For example you could have $f(n+1)=f(n)\cdot (n+1)$ and $f(3)=42$, from which you could demonstrate $f(5)=840$ and $f(0)=7$. If you start with $0!=1$ then it is in the definition; if you start at $1!=1$ then $0!=\frac{1!}1=1$ is a demonstration. $\endgroup$ – Henry Oct 29 '14 at 21:10
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    $\begingroup$ A proper definition of a function $f\colon A\to B$ should start by specifying $A$ (and $B$) $\endgroup$ – Hagen von Eitzen Oct 29 '14 at 21:14
  • $\begingroup$ @HagenvonEitzen OP implicitly defines $$n! := \begin{cases} n(n-1)! & n > 1\\ 1& n=1\end{cases}$$ with domain $\mathbb N$ as said ($0\notin\mathrm{dom}(!)$ here) $\endgroup$ – AlexR Oct 29 '14 at 21:16

Depending on the equations suitable to define $n!$, there are multiple possible answers:

If $n! := n (n-1)!, 0!=1$ is used, it's a plain definition.
If $1! = 1$ is used as the starting point, it can be proved that $0!:=1$ is a consistent extension. (this is what you did)
If $n! := \prod_{i\in \mathbb N, i \le n} i$, it follows from the definition of the empty product, $\prod_{k\in\emptyset} f(k) := 1$.
If $n! := \Gamma(n+1), n\in\mathbb N$, it follows from the definition because $\Gamma(1) = 1$.
If $n! := |S_n| = |\{f: A\to A \text{ bijective}\}|$ with $|A| = n$, there is only one set with $0$ elements, $\emptyset$ and only one function $f: \emptyset \to\emptyset$


There are multiple equivalent ways of defining the factorial function. Since the exact choice of definition doesn't matter in the long run, people don't generally make a big deal about which definition they're choosing. In some contexts, this would be a definition, whereas in others it would be a proof from the definition.

  • $\begingroup$ The questions changes then... which are the definitions taken to make this a demonstration? $\endgroup$ – hkviktor Oct 29 '14 at 21:18
  • $\begingroup$ @hkviktor There are many ways to do that. If you accept $\prod_\emptyset = 1$ I listed four variations where $0!$ is not directly defined; if you consider $\prod_\emptyset = 1$ part of the definition, there are still three examples remaining. $\endgroup$ – AlexR Oct 29 '14 at 21:21

In demonstrating that $0! = 1$, you assumed that $1! = 1$.

If you define $n!$ by

  1. $1! = 1$

  2. $n! = n(n - 1)!$ for $n \in \mathbb{N}$

with $\mathbb{N}$ defined as the set of positive integers, then you can demonstrate that $0! = 1$ by substituting $1$ for $n$.

If, as some mathematicians do, you define $\mathbb{N}$ to be the set of non-negative integers, it would make sense to define $n!$ by

  1. $0! = 1$

  2. $(n + 1)! = (n + 1)n!$ for $n \in \mathbb{N}$, with $n \geq 1$

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    $\begingroup$ Actually, if you use your first definition to define the factorial function for $n\in\mathbb N$, then $0!$ is undefined. $\endgroup$ – Jack Lee Oct 29 '14 at 21:10
  • $\begingroup$ Thanks for pointing that out. I will revise. $\endgroup$ – N. F. Taussig Oct 29 '14 at 21:11

Another way of showing why this should be true: \begin{align} 5!&=5\cdot4\cdot3\cdot2\cdot1\cdot(1)\\ 4!&=4\cdot3\cdot2\cdot1\cdot(1)\\ 3!&=3\cdot2\cdot1\cdot(1)\\ 2!&=2\cdot1\cdot(1)\\ 1!&=1\cdot(1)\\ 0!&=(1) \end{align} The first five lines are easy to verify, and the last line follows by analogy.

  • $\begingroup$ Could we not append another multiplication by one to the right of each line, and conclude that $(-1)!=1$ by this reasoning? $\endgroup$ – Peter Huxford Oct 30 '14 at 3:26
  • $\begingroup$ @Peter Um... good point. I suppose we could say that we only delete from the numbers counting down from $n$ to $1$... but now it's getting arbitrary. I don't know. $\endgroup$ – Akiva Weinberger Oct 30 '14 at 3:36
  • $\begingroup$ I think a possibly better way to see why it should be true, is that when you divide $n!$ by $n$, you get $(n-1)!$. So we would expect $0!=1!/1$. This also gives us $(-1)!$ as being infinite, which is consistent with the gamma function. However it is essentially the same reasoning as the OP. $\endgroup$ – Peter Huxford Oct 30 '14 at 3:41
  • $\begingroup$ @Peter Yeah, I was trying to be a bit original... I slightly edited my post, but I doubt it really changes anything. $\endgroup$ – Akiva Weinberger Oct 30 '14 at 3:42

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