Why is 0! equal to 1?

In this video I talk about the definition of factorial and the value of 0! #math #factorial #zero #integral

[magma]

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.

[jmle-hl]

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!

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.

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 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).

[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