Proving that the sequence $n^{(-1)^{n}}$ is divergent I would like to prove that the sequence $n^{(-1)^{n}}$ is divergent. 
My thoughts: I know $(-1)^n$ is divergent, so $n$ to the power of a divergent sequence is still divergent? I am not sure how to give a proper proof, pls help!
 A: The argument that "n to the power of a divergent sequence is divergent" does not make sense (consider $n^{-n}$ for example.). Regarding your sequence: if this sequence were convergent against some limit value $a$, then every subsequence would have to converge against the same value $a$. Now look at the subsequences for $n$ even and odd, respectively.
A: Hint. By setting $$u_n:=n^{(-1)^n}$$ one gets that
$$
\lim_{n \to \infty}u_{2n}=\infty \neq 0 =\lim_{n \to \infty}u_{2n+1}
$$ thus the sequence $\left\{ u_n \right\}$ is divergent.
A: note that $$(-1)^{n}=-1$$ if $n$ is odd and $$(-1)^{n}=1$$ if $n$ is even. 
A: Set $u_n=n^{(-1)^n}$? Explicitly, $u_{2n}=2n$,  $u_{2n+1}=\dfrac1{2n+1}$.
If the sequence were convergent, all subsequences would converge to the same limit.  However  we see the subsequence of odd terms converges to $0$, whereas the subsequence of even terms tends to $+\infty$.
A: There are lots of correct answers. Here's a suggestion for how to attack a problem like this.
Before you try to invoke abstract principles like

$n$ to the power of a divergent sequence is still divergent

which you rightly wonder about (hence your "?") try writing out the first few terms:
$$
1, 2, \frac{1}{3}, 4, \frac{1}{5}, 6, \ldots
$$
Then you can easily see that the sequence doesn't converge and can set about proving it.
A: Or to be direct...
Let $r \in \mathbb R$.  Let $\epsilon > 0$ 
For any $M$ let $m > \max (r +  \epsilon, M); m$  even. So $m - r > \epsilon > 0$.
Then $|m^{(-1)^m} - r| = |m -r|= m-r > \epsilon$.  So the sequence doesn't converge to any real $r$.
A: If $n$ even then $-1^n = 1 $
$$n^1 = n\Rightarrow \lim\limits_{n \rightarrow \infty}{({n})} = \infty$$ 
If $n$ odd then $-1^n = -1 $
$$n^{-1} = \frac{1}{n}\Rightarrow \lim\limits_{n \rightarrow \infty}{(\frac{1}{n})} = 0$$
