Understanding the proof of $\displaystyle\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$ I need help understanding this:
Prove that $$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$$

$$(\sqrt n)^\frac{1}{n}\geq 
(\sqrt1)^\frac{1}{n}=1, \forall n\in\Bbb N$$
  By binomial theorem we get for $n\geq 2$
$$n=((\sqrt n)^\frac{1}{n})^n=[1+((\sqrt n)^\frac{1}{n}-1)]^n=\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k$$
  $$\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2=1+\frac{n(n-1)}{2}(\sqrt n^\frac{1}{n}-1)^2$$
  $$\Rightarrow (\sqrt n^\frac{1}{n}-1)^2\leq \frac{2}{n}$$
  $$\Rightarrow \sqrt n^\frac{1}{n}\leq 1+\frac{\sqrt 2}{\sqrt n}$$
  $1+\frac{\sqrt 2}{\sqrt n}$ approaches $1$ for $n \rightarrow \infty$ so by the sandwich theorem we get $\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$

I don't understand this part: 
$$
\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2.$$ Could someone explain how they got that on the RHS?

I'd like to offer another method, using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$
$$n-1=(n^{\frac{1}{2n}})^{2n}-1=\\(n^{\frac{1}{2n}}-1)(n^{\frac{1}{2n}(2n-1)}+n^{\frac{1}{2n}(2n-2)}+...+n^{\frac{1}{2n}2}+n^{\frac{1}{2n}}+1)\geq$$
using AM-GM
$$\geq(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}(2n-1)}\cdot n^{\frac{1}{2n}(2n-2)}\cdot ...\cdot n^{\frac{1}{2n}2}\cdot n^{\frac{1}{2n}}}+1\right)=\\
(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}(2n-1+2n-2+...+1)}}+1\right)=
\\(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}\frac{(2n-1)2n}{2}}}+1\right)=\\
(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt{n}+1\right)$$
Also $\ln{x}\leq x-1,x>0$ which means $$\ln{n^{\frac{1}{2n}}}=\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1$$
Altogether 
$$\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1\leq \frac{n-1}{(2n-1)\sqrt{n}+1}<\frac{1}{2\sqrt{n}}$$
