Why does (n+1)! equal (n+1)n?

[CinderBlockFist]

I'm confused from one step to the next in the book.

In the first step, they have (n+1)!

then in the next step the write that as (n+1)n!


so (n+1)! = (n+1)n! ?

I don't understand that, can someone explain it to me ? ( I am very unfamiliar with factorials, this is like some of the first times I have seen it in computations.

 

[Sirus]

Let's take a general case:
$$a!=a(a-1)(a-2)(a-3)...$$
So in the problem:
$$(n+1)!=(n+1)(n+1-1)(n+1-2)(n+1-3)(n+1-4)...=(n+1)(n)(n-1)(n-2)(n-3)...$$
Now you can see that the terms after $(n+1)$ are just $n!$, according to our general definition above, so:
$$(n+1)!=(n+1)n!$$

 

[winhog]

The easiest way to see it is a concrete example I think. Take n as 5... then (n+1)! is 6! , which is 6*5*4*3*2*1. (n+1)n! is the same thing because n+1 is 6, then n! is 5*4*3*2*1, giving you 6*5*4*3*2*1.

edit: posted late I guess

 

[newcool]

n! = 1 *2 * 3 ... *n

(n + 1)! = 1 * 2 * 3 ... *n *(n+1)

Since 1*2*3...*n = n!, you can rewrite the statement so that

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

 

[CinderBlockFist]

Ohhh i get it now, thanks guys! all of your responses helped me visualize it, ty.