Prove by induction that $n^2 < 2^n$ for all $n \geq 5$? So far I have this:
First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.
Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.
So I started with $(n+1)^2 = n^2 + 2n + 1$. Because $n^2 < 2^n$ by the hypothesis, $n^2 + 2n + 1$ < $2^n + 2n + 1$. As far as I know, the only way I can get $2^{n+1}$ on the right side is to multiply it by $2$, but then I get $2^{n+1} + 4n + 2$ on the right side and don't know how to get rid of the $4n + 2$. Am I on the right track, or should I have gone a different route?

Remark that
$$2^{n+1} = 2 \cdot 2^n > 2 n^2 = n^2 +n^2> n^2+2n+1 = (n+1)^2.$$
Indeed $n^2-2n-1=(n-1)^2-2>0$ for $n \geq 5$.
