Prove for $k\in \mathbb{Z}$ that $\lim\limits_{n\to\infty}\sqrt[n]{n^k}=1$
My attempt:
Observe $k=1$, then $\lim\limits_{n\to\infty}\sqrt[n]{n}=1$. Let $x_n:=\sqrt[n]{n}-1$. Then: $$n=(1+x_n)^n>1+\begin{pmatrix}n \\ 2 \end{pmatrix}x_n^2 \Longrightarrow n-1>\frac{n(n-2)}{2}x_n^2 \Longrightarrow x_n<\sqrt{\frac{2}{n}} \Rightarrow \lim\limits_{n\to\infty}x_n=0$$
I am stuck proving it for every other $k$. Can anyone help me out?