# How to show that n! < (n+1)^n

1. Dec 7, 2011

### brookey86

1. The problem statement, all variables and given/known data

n! < (n+1)n

I am not looking for a proof, just a way to simplify this equation mathematically.

2. Relevant equations

3. The attempt at a solution

As far as I can see, I cannot simplify this any further. Is there something I can divide out of both sides, for example?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 7, 2011

### zcd

$$n! = \prod_i=1^n i$$
$$(n+1)^n =\prod_i=1^n (n+1)$$

What can you say about the relation between i and n+1?

3. Dec 7, 2011

### zcd

EDIT: bad latex
$$n! = \prod_{i=1}^n i$$
$$(n+1)^n =\prod_{i=1}^n (n+1)$$

4. Dec 7, 2011

### brookey86

I can see that i will always be less than n+1. But is there a way to compare the two without using the product summation symbol?

5. Dec 7, 2011

### zcd

You could confirm what you wrote:
$$(\frac{1}{n+1})( \frac{2}{n+1} ) ...( \frac{n-1}{n+1} )( \frac{n}{n+1} )< 1 \Rightarrow n! < (n+1)^n$$

6. Dec 7, 2011

### HallsofIvy

Staff Emeritus
Since that formula depends upon the postive integer n, you might consider proof by induction.

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