Understanding induction proof with inequalities I'm having a hard time proving inequalities with induction proofs. Is there a pattern involved in proving inequalities when it comes to induction? For example:
Prove ( for any integer $n>4$ ): 
$$2^n > n^2 \\ $$
Well, the skipping ahead to the iduction portion, here's what I've got so far:
let $n=k$, then $2^{k+1} > (k+1)^2 $. 
Starting with the LHS 
$$\begin{align}     \\  2^{k+1}=2\cdot2^k  \\  > 2\cdot k^2  \end{align}$$
And that 's where it ends for me. What would be the next step, and what suggestions might you give for future proofs like this? Do we always need to prove that the left hand side is equal to the other?

Working off Eric's answer/approach:
Let P(n) be the statement that $2^n \gt n^2$.
Basis step: (n=5) $2^5\gt 5^2 $ which is true.
Inductive step: We assume the inductive hypothesis that $P(k)$ is true for an arbitrary integer $k\ge5$.
$$ 2^{k} \gt k^2 \text{ (IH)}$$
Our goal is to show that P(k+1) is true. Let's multiply each side by two:
$$ 2*2^{k}>2(k)^2 $$
$$ 2^{k+1}>2k^2 $$
If this statement is true, then $2k^2$ $\ge$ $(k+1)^2$ must also be true when $k\ge5$:
$$ 2k^{2}\ge(k+1)^2 $$
$$ 2k^{2} \ge k^2+2k+1$$
$$k^2\ge2k+1$$
And it is. Therefore $\forall{_{k\ge5}}P(k) \implies P(k+1)$ and $\forall{_{n\ge5}}P(n)$ follows by induction. 
