Originally Posted by

**oli1230** I have an exam coming up on Wednesday and am still trying to decipher the logical problem-solving steps needed to solve a question by induction like this (from the practice questions pool):

For all natural numbers n, 3^n < (n+3)!

I have deciphered some of the problem solving into a couple base cases:

case 1: n = 0, so 3^0 < (0+3)!, 1 < 6, so the base case works but I don't know where to go from here.

case 2: n = 0 works as shown in the last example, and then the equation can be rewritten to get another base case to use induction.

The rewritten equation is 3 < ((n+3)!)^1/n, just taking the nth root of both sides. By using n = 1 as a new base case and getting 3 < 24, I have shown that this method also works. But how can I manipulate this equation to prove that it keeps working? How can I use ((n + 1) + 3)! to show that this will always be getting bigger and thus will remain bigger than 3?