1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Dec 7, 2011 #1
    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. jcsd
  3. Dec 7, 2011 #2

    zcd

    User Avatar

    [tex]n! = \prod_i=1^n i[/tex]
    [tex](n+1)^n =\prod_i=1^n (n+1)[/tex]

    What can you say about the relation between i and n+1?
     
  4. Dec 7, 2011 #3

    zcd

    User Avatar

    EDIT: bad latex
    [tex]n! = \prod_{i=1}^n i[/tex]
    [tex](n+1)^n =\prod_{i=1}^n (n+1)[/tex]
     
  5. Dec 7, 2011 #4
    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?
     
  6. Dec 7, 2011 #5

    zcd

    User Avatar

    You could confirm what you wrote:
    [tex](\frac{1}{n+1})( \frac{2}{n+1} ) ...( \frac{n-1}{n+1} )( \frac{n}{n+1} )< 1 \Rightarrow n! < (n+1)^n[/tex]
     
  7. Dec 7, 2011 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since that formula depends upon the postive integer n, you might consider proof by induction.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook