<s>[INST] 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?
 [/INST] First, we prove a simple lemma. 
$$ 
\begin{align}
n & \gt 5\\ 
\implies n-1 &\gt 4\\ 
\implies (n-1)^2 &\gt 16 \gt 2 \\
\implies n^2 - 2n + 1 &\gt  2\\ 
\implies n^2 &\gt 2n +1\\
\end{align}
$$ 
Now, we start with our induction step. 
$$
\begin{align}
n^2 &\lt 2^n \text{ From hypothesis}\\ 
\implies 2n^2 & \lt 2^{n+1}\\
\implies n^2 + n^2 &\lt 2^{n+1}\\ 
\implies n^2 + 2n + 1 &\lt 2^{n+1} (n^2 \gt 2n+1)\\ 
\implies (n+1)^2 &\lt 2^{n+1} \\
\end{align}$$
</s>