Prove : For $k\in \mathbb{Z}: \lim\limits_{n\to\infty}\sqrt[n]{n^k}=1$ Proof: For $k\in \mathbb{Z}: \lim\limits_{n\to\infty}\sqrt[n]{n^k}=1$
Proof:
Observe $k=1$. Then $\lim\limits_{n\to\infty}\sqrt[n]{n}=1$. Let $c_n:=\sqrt[n]{n}-1$, then: $$\begin{gather}
n=(1+c_n)^n>1+\begin{pmatrix}n \\ 2 \end{pmatrix}c_n^2 \Longrightarrow n-1>\frac{n(n-1)}{2}c_n^2 \Longrightarrow c_n<\sqrt{\frac{2}{n}} \Longrightarrow \lim\limits_{n\to\infty}c_n=0 \quad  \Box
\end{gather}$$
I don't understand the first step with Bernoulli's inequality - why $\begin{pmatrix}n \\ 2 \end{pmatrix}$? 
Secondly, I don't understand why $n-1>\frac{n(n-2)}{2}c_n^2$ implies $c_n<\sqrt{\frac{2}{n}}$...
Any ideas?
 A: If you expand $(1+c_n)^n$, you'll obtain $1 + {n\choose 1} c_n + {n \choose 2} c_n^2 + \ldots$, and the inequality should now be obvious.
For the second point, their proof is actually incorrect here. Note that $\frac{n(n-2)}{2}c_n^2 < n-1$ implies that $c_n^2 < \frac{2}{n} \cdot\frac{(n-1)}{(n-2)}$. You can't get what they stated here because $\frac{n-1}{n-2}$ is actually greater than $1$, but you can get something just as good; note that $\frac{n-1}{n-2} = 1 + \frac{1}{n-2} < 2$ for all $n>3$, so we get $c_n^2 < \frac{4}{n}$ for $n>3$. So we have $c_n < \sqrt{\frac{4}{n}}$, and the proof ends in the same way.
A: It is well known that
$$(x+1)^n = x^n + \binom{n}{n-1} x^{n-1}+\binom{n}{n-2}x^{n-2}+...+\binom{n}{2}x^2+\binom{n}{1}x+1$$
Since every term is positive we have $(c_n+1)^n>1 + \binom{n}{2} c_n^{2}$.
For your second question
$$n-1>\frac{n(n-2)}{2}c_n^2\Rightarrow c_n^2<\frac{2(n-1)}{n(n-2)}$$
Now there's a mistake since $n-1>n-2$ you can't conclude that $$c_n^2<\frac{2(n-1)}{n(n-2)}<\frac{2}{n}$$ See @auscrypt answer for a just as good result.
Edit: It seems like there's a much easier way to do that. Using the arithmetic of limits we have that
$$\lim_{n\rightarrow\infty} \sqrt [n]{n^k} = \lim_{n\rightarrow\infty} (\sqrt[n]{n})^k =(\lim_{n\rightarrow\infty} \sqrt[n]{n})^k= 1^k =1$$
A: Let $L=\lim_{n\to\infty}n^{\frac{k}{n}}$
Take log both sides, we get  $$logL=\lim_{n\to\infty}\frac{klogn}{n}=\lim_{n\to\infty}\frac{\frac{k}{n}}{1}=0$$ 
(using L'Hospital Rule)
So $logL=0$ which gives $L=1$
Hope this will be helpful!
