Why is $(n+3)\times(n+2)\times(n+1)! =(n+3)!?$

Question

How is $(n+3)\times(n+2)\times(n+1)! =(n+3)!?$

I know $n!= n\times(n-1)\times(n-2)\dots$. But that doesn’t seem to work in this circumstance. Please help explain

(This one too if you can). $(n+1)n!$

Answer 1

Unless I'm missing something, simply testing it with a number e.g. 2 should explain how it works very well.

let n = 2

n + 3 = 5

n + 2 = 4

(n + 1)! = 3 x 2 x 1 (the rest of the factorial of (n + 3)!)

When multiplied, you will get 5 x 4 x (3 x 2 x 1). Ending you with the definition of (n + 3)!

I'm assuming you know that (x + 1)! = (x + 1)(x)!

You are basically multiplying following two numbers to reach a factorial two higher.

Your statement can be simplified to (x + 2)(x + 1)(x)! = (x + 2)! to help understanding.

Also what do you mean by (n + 1)n! It would simplify to (n + 1)! as you are simply multiplying by the next number in the factorial expansion.

e.g.

Let n = 4

n! = 4 x 3 x 2 x 1

n + 1 = 5

n! x (n + 1) = 5 x (4 x 3 x 2 x 1)

(n + 1)! = 5 x 4 x 3 x 2 x 1

I hope this is easy to understand.

In general, (n + x)(n + (x - 1))(n + (x - 2)) ... |until x - (x - 1) is reached| ... (n)! = (n + x)!

For example, (n + 5)(n + 4)(n + 3)(n + 2)(n + 1)n! = (n + 5)!