Proving that $A=\{(-2)^n : n \in \mathbb{N} \}$ is unbounded 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!
 A: What does it mean to be bounded? $A$ is bounded if there exists some positive number $M$ such that $|a|\leq M$ for all $a\in A.$ You want to show $A$ is unbounded, so that there is no such $M$ that satisfies that condition. One way to do this is to pick an arbitrary positive number, and find something in $A$ which has larger magnitude than what you picked. If you can do that for any $M>0$ then $A$ can't be bounded. 
So to start your proof: Let $M>0.$ We can find $a\in A$ such that $|a|> M$ by picking $n$ ...
A: I would simply prove by induction that $2^n>n$ for $n\in\Bbb N$ and then note that for any $m\in\Bbb N$, $(-2)^{2m}=2^{2m}>2m>m$, so $A$ has no upper bound. If you also want to show that $A$ has no lower bound, just observe that $(-2)^{2m+1}=-2^{2m+1}<-(2m+1)<-m$. (Here I’m using the fact about odd and even powers of $-2$ that you already proved by induction.)
A: First, there exists positive integer $n$ such that $2^n > M$ whenever $M > 0$.
Supposing that $N = 1 + \lfloor M \rfloor > \max(1,M)$, we have $2^N = (1+1)^N > 1+N > 1+M > M$, so we only need to set $n = N$.
Next, we prove that there exists positive integer $n$ such that $\left|(-2)^n\right| > M$, which is equivalent to $2^n > M$, so $(-2)^n$ is unbounded. Q.E.D
A: That's essentially correct, but you should be more precise. I would try a proof by contradiction.
We are trying to show the sequence is unbounded. To do this, we will show that it cannot be bounded. Suppose the sequence was bounded, so that every terms is less than $M$ in absolute value. You need to show that for every $M\in\mathbb{N}$, there is some $n$ so that $|(-2)^n|>M$. Once you show this, you have a contradiction, so the proof is complete.
