# mathematical induction (solved but Im not sure)

• Oct 5th 2008, 11:21 AM
narbe
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 .... (Wondering)
Thanks in advance for helping me
• Oct 5th 2008, 11:53 AM
o_O
Quote:

Originally Posted by narbe
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:
$\displaystyle \begin{array}{rcl}(k+1)! & = & (k+1) \cdot k! \\ & < & (k+1) \cdot k^k \\ & = & (k+1) \cdot \underbrace{k \cdot k \cdot \hdots \cdot k}_{k \text{ times}} \\ & < & (k+1) \cdot \underbrace{(k+1) \cdot (k+1) \cdot \hdots \cdot (k+1)}_{k \text{ times}} \\ & = & \hdots \end{array}$

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:
$\displaystyle n! = n(n-1)(n-2)\hdots (2)(1) < n\underbrace{(n)(n) \hdots (n)(n)}_{(n-1) \ \text{times}} = n^n$
• Oct 5th 2008, 04:23 PM
o_O
Quote:

Originally Posted by narbe
hello my friend.

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).