Why is 0! equal to 1?

[@magma90]

I personally prefer the definition of factorials as (n+1)!=(n+1)*n!, 1!=1, because then we get 1!=1*0! => 1=1!=0! as we don’t need to define real numbers first, and it works in any set with multiplication, as 1 is defined as multiplicative identity. if we don’t have n! defined for all numbers in the set, we can always introduce additional structure, such as the gamma function.

[@alvaro.lozano-robledo]

@@magma90 The formula (n+1)! = (n+1)n! is only defined for n>=1 so you are just declaring 0!=1 as a definition anyway so that its value extends the formula.

[@alvaro.lozano-robledo]

Would you also define (-1)! = 1/0 then?

[@magma90]

@@alvaro.lozano-robledo it is actually defined for zero as we are using an implicit definition (not explicit). By algebraic manipulations we have (n+1)!/n!=n+1 or n!=(n+1)!/n+1

[@magma90]

@@alvaro.lozano-robledo as long as 1/0 is defined, then yes (-1)!=1/0. However as 1/0 makes algebra more difficult, then (-1)! is undefined because there are no real numbers “x” such that x*0=1. Also (-1)! does equal to 1/0 in the extended complex numbers where we have the complex infinity at the top of the Riemann sphere (which is 1/0)

[@pritampal3764]

⁠@@magma90 why not use gamma function?

[@jmle-h8l]

It isn't particularly illuminating, but from an algebraist's POV if we define n! to the product of all x such that 1≤x≤n, then 0! is automatically 1 because it is an empty product, and empty products are always the multiplicative identity. To fit with the combinatorial explanation, we can also define factorial as the number of bijections from the set [n] = {k∈ℕ: 1≤k≤n} to itself; then since [0] is the empty set, we have exactly one bijection - the empty map. Lots of good reasons for it to be 1!

[@MuffinsAPlenty]

"It isn't particularly illuminating" If you explain why we define the empty product as 1 (the multiplicative identity is the only thing the empty product could be if we expect to to be compatible with the generalized associative property of multiplication), then I think it's pretty illuminating. But yeah, just saying "the empty product" isn't, in and of itself, very illuminating.

An example; The product of (n+4)(n+7)/(n+2)(n+9) from n=1 to N=7 is (7+4)!(7+7)!/(7+2)!(7+9)! • 2!9!/4!7!, Where the factors without parantheses is the value for product at N=infinity. We have 11!14!/9!16! • 8•9/3•4 = 10•11/15•16 • 6 = 11/24 • 6 = 11/4. So the value of the partial product at N=7 is 11/4 and at N = infinity is 6.

[@Redstoner34526]

I like the combinatorics reason more

If and only if A+B=C+D (so that the product have a chance of being convergent (easily realised by the divergence of the harmonic series)), the infinite product (n+A)(n+B)/(n+C)(N+D) from n=1 to infinity, has the numerical value C!D!/A!B!, since its partial products has the value ((n+A)!(n+B)!/(n+C)!(n+D)!)(C!D!/A!B!). By simple arguments of continuity and by plugging in values, this obviously has the consequence that 0!=1. (It's "kind of" analogous to n^0 = 1, but with expressions rather than numbers).

In the partial product "N" might have been a better way than "n" marking up to what number the partial product is taken. Also I wrote "N" in one instance instead of "n".

[@JustDeerLol]

Here's is a simpler way n! = n(n-1)(n-2)...(2)(1) = n * (n-1)! 1! = 1 * 0! 1 = 1* 0! 0! = 1

Not really full argument in general. But at a KZbin discussion common level, it's OK. I would write this just n!=(n-1)!*n, 1!=0!*1. But if you want look at my complete elementary proof above. It requires a contiuity argument, but that argument is self-evident as Cavalieri's principle.