2ⁿ>n² for all integer n > 4

For n =5

2 ² > 5 ² or 32 >25 (True)

Assume n = k, 2 ^{k >} k ²

2^k+1 = 2 * 2^k > 2 *k² > (k+1)²

Here,

(k 1 )² > 4² > 2

K² 2k +1 > 2

K²-2k-1>0

2k²-2k-1>k²

2k²>k²+2k+1

2k²>(k+1)²