I am trying to prove that $A=\{(-2)^n : n \in \mathbb{N} \}$ is unbounded.
What I did was first to show that for every $n \in \mathbb{N}$ if $n$ is even then $(-2)^n = 2^n$ and if $n$ is odd then $(-2)^n = -2^n$ (I did it by induction on $n$).
Then I show that for every $n \in \mathbb{N}$ there exist $k \in \mathbb{N}$ such that $(-2)^n \lt (-2)^k$ because if $n$ is even then $(-2)^n=2^n\lt 2^k$ Where the last inequality holds by the definition of natural powers. The case where $n$ is odd is obvious now.
Am I right? Is the a more elegant way of proving so? thanks!