1. The problem statement, all variables and given/known data Original Problem Text: http://img264.imageshack.us/img264/5882/problem3pn7.th.gif [Broken] Prove that, if n [tex]\in[/tex] [tex]N[/tex] and n [tex]\geq[/tex] 2, then 2^(n^2) > n^(n). 2. Relevant equations Some Hints: *** (n-1)^(n-1) * 2^(n-1) = (2n - 2)^(n-1) *** If n >= 2 then (2n - 2) >= n. *** If n >= 1 then 2^(n) >= n. 3. The attempt at a solution Using Induction: For the base step, suppose that n = 2. This implies that 2^(2^2) = 16, and 2^(2)= 4, and 16 > 4, so the base step is proven true. Suppose that n>2, and that the theorem has been proven for all smaller values of n. Then we compute: 2^(n^2) > n^n 2^(n-1)^2 > (n-1)^(n-1) ***Multiply both sides by 2^(n-1)*** 2^(n-1)^2 * 2^(n-1) > (n-1)^(n-1) * 2^(n-1) 2^(n-1)^2 * 2^(n-1) = 2^(n^2 - n) = 2^(n(n-1)) 2^(n(n-1)) > (2n -2)^(n-1) ***Now here's the screwy part - I took the (n-1) root across the inequality. I have my doubts about the mathematical legality of such an operation. *** 2^(n) > 2n - 2 2n -2 [tex]\geq[/tex] n (since n [tex]\geq[/tex] 2) 2^(n) > n 2^(n^2) > n^(2) Does this proof make any sense at all? Does it even remotely resemble a proof? I should point out that I'm using (n-1) in place of (n+1) because that's the way my professor did it. Thanks a lot for any help!