# SOLUTION: Prove n!>n^2 for n>=4 and n!>n^3 for n>=6 by using Principle of Mathematical Induction.

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 Question 633231: Prove n!>n^2 for n>=4 and n!>n^3 for n>=6 by using Principle of Mathematical Induction.Answer by Edwin McCravy(8909)   (Show Source): You can put this solution on YOUR website!Prove n! > n² for n>=4 ```We begin by showing that it is true for the smallest possible value of n, which is 4: 4! > 4² 24 > 16 which is true. Assume for n <= k (1) k! > k² <--- we are assuming this. We need to show that assumption (1) leads to (2) (k + 1)! > (k + 1)² <--- we do not know this yet! Since (k + 1)² = k² + 2k + 1, the 2k + 1 tells us that we should add 2k + 1 to both sides of assumed inequality (1): (3) k! + 2k + 1 > k² + 2k + 1 = (k + 1)² <--- we know this Now we just need to show that the left side of (3) is less than the left side of (2) That is, we need to show that (4) k! + 2k + 1 < (k + 1)! <--- we do not know this yet which is equivalent to (5) k! + 2k + 1 < (k + 1)k! <--- we do not know this yet which is equivalent to (6) k! + 2k + 1 < k·k! + k! <--- we do not know this yet which is equivalent to (7) 2k + 1 < k·k! <--- we do not know this yet Which is equivalent to the following (upon dividing through by k): (9) 2 + < k! <--- we DO know this. We DO know (9) is true. Here's why: The left side is always less than 3 and since k >= 4, the right side k! is always 4! = 24 or greater so we do know (9) is true. Now since all those inequalities are equivalent we can reverse the steps (9) 2 + < k! is true, therefore, multiplying thru by k, (8) 2k + 1 < k·k! is true, therefore adding k! to both sides: (7) k! + 2k + 1 < k·k! + k! is true, therefore factoring the right: (6) k! + 2k + 1 < k!(k + 1) is true, and the right side is (k + 1)!, so (5) k! + 2k + 1 < (k + 1)k! is true. So: (4) k! + 2k + 1 < (k + 1)! is true, which is the same as (k + 1)! > k! + 2k + 1 and since (3) is true, which is (3) k! + 2k + 1 > k² + 2k + 1 = (k + 1)², we have (k + 1)! > k! + 2k + 1 > k² + 2k + 1 > (k + 1)² which is what we needed to prove. The proof of the other one is similar. It amounts to taking the inequality you have to prove, simplifying it using equivalent inequalities to get something that you know is true, then you reverse the steps. Edwin```