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

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Homework Help Overview

The discussion revolves around the inequality n! < (n+1)^n, with participants exploring ways to simplify or analyze this mathematical expression without seeking a formal proof.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the factorial n! and the expression (n+1)^n, questioning whether simplifications can be made. Some consider the implications of the product representations of both sides and the relative sizes of the terms involved.

Discussion Status

There is an ongoing exploration of the inequality, with some participants suggesting ways to confirm the relationship through manipulation of the terms. Others propose considering proof techniques such as induction, indicating a productive direction in the discussion.

Contextual Notes

Participants are operating under the constraint of not providing a complete proof, focusing instead on simplification and analysis of the inequality.

brookey86
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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 positive integer n, you might consider proof by induction.
 

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