TITLE: $\lim\limits_{n\to\infty} \sup \sqrt[n]{n^k}$ for $k \in \mathbb{N}$?
QUESTION [5 upvotes]: What is $$\lim\limits_{n\to\infty} \sup \sqrt[n]{n^k} \ \ \ \text{where } k \in \mathbb{N}\text ?$$ I've seen the proof that $\lim\limits_{n\to\infty} \sqrt[n]n = 1$. I believe the answer is $1$ for all $k$ because $\sqrt[n]n$ approaches $1$, so a sequence that approaches $1$ raised to an exponent, I believe, should also approach $1$. 
I don't know how to prove this, mainly because of the $\lim\limits_{n\to\infty}\sup$ concept. I learned $\lim\limits_{n\to\infty}\sup s_n$ as the $\sup$ of all subsequential limits of $s_n$. Intuitively I know it's the $\inf$ of all $\sup$s of the sequence as $n\to\infty$. But I don't know how to use that definition /intuition to solve problems like this.

REPLY [3 votes]: Observe that $$\sqrt[n]{n^k}=(\sqrt[n]{n})^k.$$ Since $f(x)=x^k$  is continuous, $f(\sqrt[n]{n})\rightarrow f(1)=1$ as $\sqrt[n]{n} \rightarrow 1$ an the result follows.