I have to prove by induction that for n>= 4 that 2^n > n^2. So i start with the base case and I get 16 >= 16 which is true. Then I assume for k that 2^k >= k^2 for k. Now I have to that 2^(k+1) >= (k+1)^2 Now going back to 2^k >= k^2 if I multiply both sides by 2 I get 2*2^k >= 2*k^2 2^(k+1) >= 2*k^2 and from here I get stuck. Any help would be appreciated. Thanks.