Are these factorial statements accurate?

[Dannbr]

so,

6! = 6*5*4*3*2*1

(n+2)! = (n+2)(n+1)(n)!

(2n+2)! = (2n+2)(2n+1)(2n)!

(500n+3)! = (500n+3)(500n+2)(500n+1)(500n)!

Are all these statements correct?

 

[Mark44]

so,

6! = 6*5*4*3*2*1

(n+2)! = (n+2)(n+1)(n)!

(2n+2)! = (2n+2)(2n+1)(2n)!

(500n+3)! = (500n+3)(500n+2)(500n+1)(500n)!

Are all these statements correct?

Yes.

 

[gb7nash]

so,

6! = 6*5*4*3*2*1

(n+2)! = (n+2)(n+1)(n)!

(2n+2)! = (2n+2)(2n+1)(2n)!

(500n+3)! = (500n+3)(500n+2)(500n+1)(500n)!

Are all these statements correct?

Yes. Using these properties can allow you to simplify something like:

\frac{(n+2)!}{n!} = (n+1)(n+2)

 

[TylerH]

You could generalize to get \frac{(n+a)!}{n!}=\prod_{i=1}^a n+i