1. The problem statement, all variables and given/known data I am asked to prove: 2n < (n+1)! , where n≥2 3. The attempt at a solution Base step: set n=2, then test 22 < (2+1)! 22 = 4 (2+1)!= 3! = 3(2)(1) = 6 so 4 < 6 , which is true. Induction hypothesis is 2k < (k+1)! Using this, prove 2(k+1) < [(k+1)+1]! = (k+2)! Attempt to solve: starting with what I know: 2k < (k+1)! Multiplying both sides by 2: 2(2k) = 2(k+1) < 2(k+1)! I know that 2(k+1)! < (k+2)! since (k+2)! = (k+2)(k+1)! and because k≥2, (k+2) will be greater than 2. Thus, multiplying (k+1)! by 2 on the LHS is less than multiplying (k+1)! by (k+2) on the RHS. Thus, since 2(k+1) < 2(k+1)! is true, then 2k+1 < [(k+1)+1]!. P(k+1) follows from P(k), completing the induction step. By mathematical induction, P(n) is true for n≥2. Thanks for any help! EDIT: fixed a couple of type-o's.