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I know that we define $n!$ as

$$n! = n\cdot(n-1)!,$$

so that $0! = 1$ follows from $1! = 1$.

However, what I would like to find out is the mathematical intuition behind $0! = 1$, if there is any.

Jose Arnaldo Bebita Dris
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  • See this question. Essentially, you want $n!=\frac {(n+1)!}{(n+1)} $. Set n=0 and see what happens. – MM8 Jan 30 '17 at 15:04
  • @DietrichBurde and that one is a duplicate of http://math.stackexchange.com/questions/20969/prove-0-1-from-first-principles – Ethan Bolker Jan 30 '17 at 15:05
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    @EthanBolker Yes, I know. But seeing that a question is already a duplicate of a duplicate makes it perhaps more convincing that the question has been sufficiently discussed on MSE. – Dietrich Burde Jan 30 '17 at 15:07
  • @TimonG. : I think OP already understands that (he stated as much), but wants some intuitive explanation in addition to the technical one. – MPW Jan 30 '17 at 15:08
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    What is a more mathematically intuitive explanation than "it follows by plugging in values into the recursive definition of the factorial"? – MM8 Jan 30 '17 at 15:09
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    Interestingly, the fact that the product of no factors is naturally defined to be $1$ appears to be less obvious than the sum of no term being defined to be $0$. But taking the logarithm explains it easily. –  Jan 30 '17 at 15:12
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    Empty products. – Simply Beautiful Art Jan 30 '17 at 15:22
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    $6$ ways to arrange three objects:$$\begin{matrix}(1,2,3)&(1,3,2)\(2,1,3)&( 2,3,1)\(3,1,2)&(3,2,1)\end{matrix}$$ $2$ ways to arrange two objects:$$\begin{matrix}(1,2)&(2,1)\end{matrix}$$ $1$ way to arrange one object:$$(1)$$ $1$ way to arrange zero objects:$$()$$ – Akiva Weinberger Jan 30 '17 at 15:43

1 Answers1

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How many ways are there to arrange zero objects in a line? Only one way, the way that arranges no objects.

Sean Roberson
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