What is the inductive definition of $n!$?

Question

Can someone explain the meaning of inductive definition of $n!$?

Answer 1

For $n\in\mathbb N_0$, we define $0!:=1$ and for $n\geq1$, we set $$n!:=(n-1)!\cdot n.$$

Inductive means that we use the value of $(n-1)!$ in order to define the next factorial, which is $n!$.

Answer 2

$\Gamma (n)=n\Gamma (n-1)\,, n\gt0$ and $\Gamma (0)=1$.

So, $\Gamma (n)$ is defined in terms of $n$ and $\Gamma (n-1)$.

Answer 3

"Induction", in general, depends upon a statement, depending on a positive integer, n, being true for n= 1 (or 0) and then, if the statement is true for n= k, it is also true for n= k+ 1.

"Defining" something that depends on a positive integer, A(n), "inductively" means defining it for n= 1 (or 0)and then defining A(n+ 1) in terms of A(n).

"n factorial", n!, is define "inductively" by defining 0! to be 1 and then defining (n+1)! to be n(n!).

For example, to find 6! using that definition, I would say that 6!= 6(5!). Of course to use that I need to first find 5!. 5!= 5(4!). 4!= 4(3!). 3!= 3(2!). 2!= 2(1!). 1!= 1(0!). 0!= 1 (by the first part of the definition) so 1!= 1(1)= 1, 2!= 2(1)= 2, 3!= 3(2)= 6, 4!= 4(6)= 24, 5!= 5(24)= 120, and 6!= 6(120)= 720.