# Thread: mathematical induction (solved but Im not sure)

1. ## 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

2. 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:
$\begin{array}{rcl}(k+1)! & = & (k+1) \cdot k! \\ & < & (k+1) \cdot k^k \\ & = & (k+1) \cdot \underbrace{k \cdot k \cdot \hdots \cdot k}_{k \text{ times}} \\ & < & (k+1) \cdot \underbrace{(k+1) \cdot (k+1) \cdot \hdots \cdot (k+1)}_{k \text{ times}} \\ & = & \hdots \end{array}$

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:
$n! = n(n-1)(n-2)\hdots (2)(1) < n\underbrace{(n)(n) \hdots (n)(n)}_{(n-1) \ \text{times}} = n^n$

3. Originally Posted by narbe
hello my friend.