In a YouTube video, James Grime of NumberPhile makes the claim that the meaning of the factorial is \[n! = \prod_{i=1}^n i\] for n > 0, and proceeds to explain why 0! = 1 using a recursive proof. This echoes what Wolfram Mathworld has to say on the subject: “The factorial n! is defined for a positive integer n as \(n! \equiv n (n-1) .. 2 \cdot 1\).”

I used to hold a similar perspective, which requires an explanation about 0! concerning the nature of mathematical convention and convenience. In my current opinion, however, the real sticking point is about the nature of definitions.

It is true that \(n! = \prod_{i=1}^n i\) for all values in the domain of the factorial function except 0. However, the (near) equivalence of two functions is not the same thing as the definitional equality of those two functions. That is to say, just because f(x) = g(x) for (nearly) all x, it’s not the case that f(x) *is defined as* g(x) or vice versa.

The original purpose of looking into factorials was examining probabilities and games of chance. The original relevance of n! was the number of distinct permutations of n different items. That is the definition of factorial: n! is the number of distinct permutations of n different items. Under that definition, 0! is not confusing at all: There is only one way that we can sort 0 items, just as there is only one way that we can sort 1 item.

Stating that the values of f(x) = n! and g(x) = \(\prod_{i=1}^n i\) happen to be the same for \(n \in \mathbb{N}^*\) is different than stating that f(x) is defined as g(x). Students already struggle with interpreting \(x = y\) as “the value of the expression x is equivalent to the value of the expression y”; confusing “is equivalent to” with “is defined as” (or with just plain “is”) doesn’t help matters.

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