Factorial Q: How to Get (n+1)!

[askor]

How do you get

(n + 1)! = (n + 1)(n)(n - 1)(n - 2) ... 3 ⋅ 2 ⋅ 1

?

Isn't (n + 1)! = (n + 1) ⋅ (n + 1 - 1) ⋅ (n + 1 - 2) ⋅ (n + 1 - 3) ⋅ (n + 1 - 4) ... and so on?

 

[Simon Bridge]

How do you get

(n + 1)! = (n + 1)(n)(n - 1)(n - 2) ... 3 ⋅ 2 ⋅ 1

?

Isn't (n + 1)! = (n + 1) ⋅ (n + 1 - 1) ⋅ (n + 1 - 2) ⋅ (n + 1 - 3) ⋅ (n + 1 - 4) ... and so on?

You need to realize that:
n+1-1 = n
n+1-2 = n-1
n+1-3 = n-2
etc

 

[jedishrfu]

or more compactly:

(n+1)! = (n+1) * (n+1-1)! = (n+1) * n!