For every integer $n \geq 1$, prove that $3^n \geq n^2$. It's been a while since I've done induction, and I feel like I'm missing something really simple. What I have is this:

Base Case: $n=1$
   $$3^n \geq n^2 \implies 3 \geq 1$$
Inductive Hypothesis 
For all integers $1 \leq n < n+1$:
  $$3^n \ge n^2$$
Inductive Step 
$$3^{n+1} \geq \left( n+1\right) ^2 \implies 3\cdot 3^n \geq n^2+2n+2$$

This is as far as I got. Is it acceptable to say $2\cdot 3^n \geq 2n+2$ and then prove that? Or is there an easier way?

The claim is clear if $n=1$.
If $n\ge 2$, by the binomial theorem, we have
$$
3^n = (1+2)^n = 1 + 2\binom n1 + 2^2\binom n2 + \dots > 2\binom n1 + 2^2\binom n2 = 2n + 2n(n-1) = 2n^2 > n^2
$$
