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?

Since $n^{1/n}\ge 1$, then all of the terms in the binomial expansion
$$\left(1+(\sqrt[n]{n}-1)\right)^n=\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k$$
are non-negative.  Therefore, we have for any $m\le n$
$$\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k\ge \sum_{k=0}^m\binom{n}{k}(\sqrt[n]{n}-1)^k$$
Taking $m=2$ and $n\ge 2$ yields
$$\begin{align}
\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k&\ge \sum_{k=0}^2\binom{n}{k}(\sqrt[n]{n}-1)^k\\\\
&=1+n(\sqrt[n]{n}-1)+\frac{n(n-1)}{2}(\sqrt[n]{n}-1)^2\\\\
&\ge 1+\frac{n(n-1)}{2}(\sqrt[n]{n}-1)^2
\end{align}$$
where the last inequality is true since $n(\sqrt[n]{n}-1)\ge 0$.
