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Math Help - mathematical induction (solved but Im not sure)

  1. #1
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    mathematical induction (solved but Im not sure)

    hello people!
    in this question it says prove the following using mathematical induction:

    n! < n^n

    p(2) = 2! < 2^2 (True)
    p(n) = n! < n^n
    p(n+1) = (n+1)! < n ^ (n+1)

    proof:

    n . n! < n^n . n
    n. n! < n ^ (n+1)
    (n+1).n! < n^ (n+1)
    (n+1)! < n ^ (n+1)

    is this true??? I dont think so ....
    Thanks in advance for helping me
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  2. #2
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    Quote Originally Posted by narbe View Post
    proof:

    n . n! < n^n . n
    n. n! < n ^ (n+1)
    (n+1).n! < n^ (n+1) How'd you get here?
    (n+1)! < n ^ (n+1)
    Consider:
    \begin{array}{rcl}(k+1)! & = & (k+1) \cdot k! \\ & < & (k+1) \cdot k^k \\ & = & (k+1) \cdot \underbrace{k \cdot k \cdot \hdots \cdot k}_{k \text{ times}} \\ & < & (k+1) \cdot \underbrace{(k+1) \cdot (k+1) \cdot \hdots \cdot (k+1)}_{k \text{ times}} \\ & = & \hdots \end{array}

    Multiplying a number by (k+1) is bigger than multiplying it by k and that is basically the argument here. You don't even need induction for this problem really:
    n! = n(n-1)(n-2)\hdots (2)(1) < n\underbrace{(n)(n) \hdots (n)(n)}_{(n-1) \ \text{times}} = n^n
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  3. #3
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    Quote Originally Posted by narbe
    hello my friend.
    about my question :

    http://www.mathhelpforum.com/math-he...-not-sure.html

    teacher want us to prove it with mathematical induction. I understand your point and thank you for your explanation. the question is now quite clear for me. but can you tell me please, how to write it as an mathematical induction prove?
    What I did was using mathematical induction. I simply gave you the inductive step. You did the base case already for p(2).
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