Note first that:


  if n = 0, then 02 = 0 and 0! = 1.
  if n = 1, then 12 = 1 and 1! = 1.
  if n = 2, then 22 = 4 and 2! = 2.
  if n = 3, then 32 = 9 and 3! = 6.


We prove by induction on n that   ≤ n!  for all  n ≥ 4.


Basis step : 

     = 16 and 4! = 24


Inductive hypothesis : 

    Assume for some integer k ≥ 4 that   ≤ k!


Inductive step :


       (k + 1)! = (k + 1)k! 

     ≥ 

     = 

     ≥ 

     = 

     ≥ 

     ≥ 

     = .


According to the method of Mathematical induction, the proof is completed.

Base case:  n=4:   and  ,  so the base case holds.



Hypothesis:  Assume  for ,   (*)



Step case:  Let n=k+1:



        <=(?)   

(where (?) is provided to show that we need to resolve this inequality)



        <=(?)   



divide both sides by k+1:

        <=(?)   



Since , and we have   by (*), we can

write  therefore   and that completes the proof.

 

RELATED QUESTIONS

prove that n^2 ≤ n! for all n ≥ 4 using mathematical... (answered by ikleyn)

Prove n!>n^2 for n>=4 and n!>n^3 for n>=6 by using Principle of Mathematical... (answered by Edwin McCravy)

Prove using mathematical induction that for all n ≥ 1,
1 + 4 + 7 + · · · + (3n −... (answered by greenestamps)

Pls help
USE MATHEMATICAL INDUCTION TO PROVE THAT 

(n+1)^n < 2n^2 for all natural... (answered by ikleyn)

Mathematical induction


How can we prove that :
(1 + 1 / 3) (1 + 5 / 4)(1 + 7 /... (answered by ikleyn)

Induction Sum:
Prove using mathematical induction that:

1/1*2*3 + 1/2*3*4 + 1/3*4*5... (answered by robertb)

use mathematical induction to prove that 
1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6... (answered by solver91311)

Use mathematical induction to prove each statement is true for all positive integers n:... (answered by math_helper)

Principle of Mathematical Induction: For all natural numbers n, n^2 <... (answered by Alan3354)