Use mathematical induction to prove that 2n > n2 , for n > 5 .

Using mathematical induction let us prove that $2^n > n^2 ,$ for $n > 5 .$

Let $n=6.$ Then $2^6=64>36=6^2,$ and hence for $n=6$ the statement is true.

Suppose that the statement is true for $n=k>5,$ that is $2^k > k^2.$

Let us prove this statement for $n=k+1.$

For this firstly let us show that $k^2>2k+1$ for $k>5.$ In this case $k-1>4,$ and hence $(k-1)^2>16.$ It follows that $k^2-2k+1>16,$ and thus $k^2-2k>15>1.$ Therefore, $k^2>2k+1.$

Taking into acount that $2^k > k^2,$ we conclude that

$2^{k+1}=2\cdot2^k>2k^2=k^2+k^2>k^2+2k+1=(k+1)^2,$

and the statement is true for $n=k+1.$

We conclue that the statement $2^n > n^2$ is true for all natural numbers $n > 5 .$