Is n! Less Than (n+1)^n Simplifiable?

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Homework Statement



n! < (n+1)n

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

Homework Equations





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?
 
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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?
 
zcd said:
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?

EDIT: bad latex
n! = \prod_{i=1}^n i
(n+1)^n =\prod_{i=1}^n (n+1)
 
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?
 
You could confirm what you wrote:
(\frac{1}{n+1})( \frac{2}{n+1} ) ...( \frac{n-1}{n+1} )( \frac{n}{n+1} )&lt; 1 \Rightarrow n! &lt; (n+1)^n
 
Since that formula depends upon the postive integer n, you might consider proof by induction.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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