應該係好容易的一題數...（不過我未諗到）
prove 2^n > n^2  for n>4 and n is an integer

when n = 5
2^5 > 5^2v , P(5) is true

assume P(k) is true, i.e. 2^k > k^2
consider n = k+1
2^(k+1) > 2*k^2 = (k+1)^2 + k^2 - 2k - 1 = (k+1)^2 + (k-1)^2  - 2 > (k+1)^2
since k > 4, (k-1)^2 - 2 > 0, therefore (k+1)^2 + [(k-1)^2 - 2] > (k+1)^2

QED