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?

There are, of course, many ways to force $2^{n + 1}$ to show up on the RHS. I like to bound the lower order terms as multiples of the leading term by repeatedly applying the given inequality $n \geq 5$ so that I can apply the inductive hypothesis in one shot. Indeed, observe that:
\begin{align*}
(n + 1)^2
&= n^2 + 2n + 1 \\
&< n^2 + 4n + 5 \\
&\leq n^2 + 4n + n &\text{since } n \geq 5 \\
&= n^2 + 5n \\
&\leq n^2 + (n)n &\text{since } n \geq 5 \\
&= 2(n^2) \\
&< 2(2^n) &\text{by the inductive hypothesis} \\
&= 2^{n + 1}
\end{align*}
as desired.
