SOLUTION: prove that n^2 ≤ n! for all n ≥ 4 using mathematical induction

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.