I need to prove that $n^2 < 2^n$ for all natural numbers $n$ greater than $4$.
I understand that you start by proving the base case of $n = 5$ and then prove the inequality substituting the inductive hypothesis for $n + 1$, but I am unsure about how to do this.
$$ 2^{x+1} = 2^x2 > 2x^2 = x^2 + x^2 > x^2 + 2x + 1 = (x + 1)^2 $$
Second inequality from the left follows from the induction hypothesis. And we get desired result if we can show $x^2 > 2x + 1$ for $x > 4$. but That is easy.
Putting in a couple extra steps for clarity, once $n\gt4$, we have
$$\begin{align} 2^{n+1}&=2\cdot2^n\cr &\gt2\cdot n^2\quad\text{ (by the induction hypothesis)}\cr &=n^2+n^2\cr &\gt n^2+n\cdot3\quad\text{ (since $n\gt3$)}\cr &\gt n^2+n\left(2+{1\over n}\right)\quad\text{ (since $n\gt1$)}\cr &=n^2+2n+1\cr &=(n+1)^2 \end{align}$$
