Proving that $2^{2n}-n^2+3^n = \Omega (2^{2n})$

Question

I need to prove that: $2^{2n}-n^2+3^n = \Omega (2^{2n})$

I started and got to this: $2^{2n}-n^2+3^n \geq 2^{2n}\cdot 3 \geq 2^{2n}\cdot 2 = 2^{2n+1}$ for every $n > n_{0} = 1$
How should I continue from here?

Answer 1

Assume $n \in \mathbb{R}$ and $n > 0$.

$$2^{2n} - n^2 + 3^n > 2^{2n}$$ $$3^n > n^2$$

This should be relatively easy to show right?