Reasons Why \(0! = 1\)

Yesterday my wife asked me why \(0! = 1\) and I realized that I off the top of my head I only had two answers: (1) it's defined that way, and (2)  the Gamma function at one is one. Well, as you might imagine, (1) was unsatisfying to her and (2) involved esoteric math that only nerds want to know about.  So I began to wonder what elementary arguments there are that \(0! = 1\). After a few minutes on google, I decided to record them here. Note that I'm including everything I can think of for completeness. Some arguments directly or indirectly reduce to a convention, others are conceptual; some are harder, some are easier; and so on.

1. Argument from definition: we define \(0! = 1\).

2. Argument from empty product convention: we define \(n! = \prod_{k = 1}^n k\) then note that the this implies that \(0!\) is the empty product, which is 1 by convention. (See footnote)

3. Argument from recursive definition: we note that \(n! = \frac{(n+1)!}{n+1}\) and put \(n = 0\).

4. First argument from combinatorics: \(n!\) is the number of ways to arrange \(n\) things and there is exactly one way to arrange zero things: do nothing.

5. Second argument from combinatorics: \(_nC_p\) where \(n > m\) is the number of ways to arrange \(n\) things into groups of \(m\) things and is computed by the formula \[_nC_p = \frac{n!}{p!(n-p)!} .\]There is only one way to arrange \(n\) things into a group of \(n\) things. Thus, for the formula to work in this case, it follows that \(\frac{n!}{n!0!}= 1 \), so \(0! = 1\).

6. Argument from the Gamma function: Let \(\text{Re}(z) >0\) and define \(\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\). Note that, for all integers \(n > 1\), \(\Gamma(n) = (n-1)!\). Next, for all \(z\) in the right complex plane, define \(z! = \Gamma(z+1)\). Since \(\Gamma(1) = 1\) it follows that \(0! = 1\).

7. Argument from bijective mappings: this is really just a more esoteric rephrasing of argument 4. Define \(n!\) as the number of bijective mappings between two sets with \(n\) elements each. For \(n = 0\) this is the number of bijective mappings from the empty set to itself. There is exactly one such mapping and therefore \(0! = 1\).

Footnote: why is the empty product equal to one? Let \(A\) be a finite set and define \(P_A = \prod_{a\in A}a\). It follows that \(\ln P_A = \sum_{a\in A}a\). Now, let \(A\) be the empty set. By convention, it follows that \(\sum_{a\in A} = 0\), thus \(\ln P_A = 0\), hence \(P_A = 1\).