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    @StephenDuffy there is an explanation for that, actually. n! can be represented as ((n+1)!) / (n+1), so 3! would be 4! / 4. It makes sense 4! = 4 * 3 * 2 * 1 and if you divide by 4 you're left with 3 * 2 * 1 which is 3!. It's also a mutually recursive factorial definition which is fun. At any rate, if you look at 0! that would be 1! / 1 which gives you 1. There you go. Another way to explain it is permutations - n! is the number of ways to arrange n items. And with zero items there is exactly one arrangement - an empty one. Commented Apr 11, 2020 at 21:25
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    @StephenDuffy but probably the most accurate explanation for 0! = 1 is also very disappointing. It's as follows: because mathematicians defined it that way. As I said, disappointing. But mathematicians do tend to define things that don't make complete sense in other terms because it then helps out with solving some formulas. Apparently, the number 1 is not prime, because if it was, you'd make some equation that uses primes a lot harder. Hence why 1 is excluded. Commented Apr 11, 2020 at 21:29
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    @VLAZ: yes, because it's defined that way. But its not arbitrary. As you pointed out, they're defined that way because they are much more useful definitions, both the definition of 0! and the definition of "prime number". More difficult ones are things like 0 ^ 0, which has several competing possibilities; yes there is a best one (0 ^ 0 = 1), and so it's often the accepted one, but there is another reasonable option (0 ^ 0 = 0), so it's much more ambiguous. Commented Apr 12, 2020 at 15:16