# Basics of Factorials

Having some trouble with some basics of factorials. Just wondering how the following factorial simplifies.

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

Shouldn't $(n+1)! = (n+1)(n)$ why is it $n!$ instead of just $n$

• Usually by definition $0!=1$ and $(n+1)!=(n+1)(n!)$ for $n \geq 0$. Commented Jan 25, 2017 at 21:32
• $(n+1)!=(n+1) \times n \times (n-1) \times ... 2 \times 1$
– Karl
Commented Jan 25, 2017 at 21:34

The definition is $n!=1\times2\times\cdots\times n$.
So $(n+1)!=1\times2\times\cdots\times n\times(n+1)=n!\times(n+1)$.

$$\frac{n}{(n+1)!}=\frac{n\cdot 1}{(n+1)(n!)}=\frac{n}{(n+1)}\cdot \frac{1}{n!}$$

remember that, by definition, $(n+1)!=(n+1)\cdot n!$

The factorial identity of interest is

$n! = n(n-1)!$

or equivalently and more useful in your case,

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

Note that the right hand side still contains a factorial. This is because of how a factorial is defined: the product of every integer from the indicated number down to 1:

$(n+1)! = (n+1)(n)(n-1)(n-2)...(2)(1)$

So $4! = 4×3×2×1$; well, $3×2×1 = 3!$, so we could write $4!=4×3!$. Or in general for any number $n$ instead of $4$, we have the very first identity listed above.

Another way to think about it is $n!$ has $n$ factors (counting 1). If $n!=n(n-1)$, well that's just two factors and not in accordance with the definition of the factorial function.

Now if you really want to have some fun with the factorial, imagine trying to extend it to non-integer numbers... Euler's Gamma function

The definition of factorial for non-negative integers is this:

$0! = 1$.

If $n \ge 1$ then $n! = n\cdot (n-1)!$.

As a consequence, $n! =n(n-1)(n-2)...2\cdot1$.

Therefore, in your question, $(n+1)! = (n+1)\cdot n!$.

I'm not sure where you got the idea that $(n+1)! = (n+1)n$.

The factorial is defined for non-negative integers in this way:

$$n!=n\cdot (n-1)\cdot (n-2)\cdots2\cdot 1$$

Therefore, $$(n+1)! = (n+1)\cdot \underbrace{n\cdot (n-1)\cdot (n-2)\cdots2\cdot 1}_{n!}\\ =(n+1)\cdot n!$$

To be clear $n!$ is $n\cdot (n-1) \cdot (n-2) \ldots \cdot 2 \cdot 1$. Likewise when we add a $1$ we get $(n+1)\cdot((n+1)-1)\cdot((n+1)-2)\cdot\ldots\cdot2\cdot 1$.

What we see often is that it is useful to factor out one of the terms in the product of a factorial, namely the first term. But we must still keep $n!$ since it is the product of the remaining terms.

So for instance $$(n+1)!=(n+1)n!$$ and depending on the size of $n$ we could possibly continue in this manner, for instance $$(n+1)!=(n+1)n!=(n+1)(n)(n-1)!$$