When $n=1$, we have $2^1 + 2 = 4 > 1^2$. Hence the claim holds. Assuming when $n=k\ge 1$, it holds that $2^k+2>k^2$. We are going to show that $2^{k+1}+2 > (k+1)^2$ will hold under this assumption. \begin{align*} & 2^{k+1} + 2\\ =& 2\times 2^k + 2 \\ >& 2\times (k^2 - 2) +2 &\qquad\scriptsize{(\because 2^k+2>k^2 \implies 2^k > k^2-2)}\\ =& 2k^2 -2 \\ =& k^2 + (k^2 -3 ) + 1\\ \ge & k^2 +2k + 1 &\qquad\scriptsize{(\because k \ge 1 \implies k^2 + 2k \ge 3\implies k^2-3 \ge 2k)}\\ =& (k+1)^2 \end{align*} Hence, by the principle of the mathematical induction, this claim holds.