Limit of $\sqrt[n]{n^k}$ for k $\in \mathbb{N}$ for $n \rightarrow \infty$

Question

So the title basically says what my question is I'm looking for a proof for the limit of the given sequence (title).

Answer 1

$$\lim_{n\to\infty} \sqrt[n]{n}=\lim_{n\to\infty}\exp{\left(\frac{\ln{(n)}}{n}\right)}=\exp{\left(\lim_{n\to\infty}\frac{\ln{(n)}}{n}\right)}=\exp{\left(\lim_{n\to\infty}\frac{1}{n}\right)}=\exp{(0)}=1$$ Where I have used L'Hôpital's rule to evaluate $\lim_{n\to\infty}\frac{\ln{(n)}}{n}$. Then due to the continuity of $f(x)=x^k$ where $k\in\mathbb{N}$. The limit is just $$\lim_{n\to\infty} \sqrt[n]{n^k}=\left(\lim_{n\to\infty} \sqrt[n]{n}\right)^k=1^k=1$$