asymptotic notations : if $0<a<b$ then $n^b=\Omega(n^a)$

Question

If $0<a<b$ then $n^b=\Omega(n^a)$.

I have learned about this quiet recently and have come across this equation. I am having difficulty proving this. Any help would be appreciated.

Answer 1

Note that $\frac{n^b}{n^a} = n^{b-a} \to \infty$ as $n \to \infty$ since $b-a>0$. That means there exists a $N$ such that for all $n \geq N$, $n^b \geq n^a$, proving the statement.