Understanding the proof of $\displaystyle\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$

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 $n(\sqrt[n]{n}-1)\ge 0$.

They are truncating the binomial expansion, discarding positive terms: if $$ r=r_0+r_1+r_2+\dots+r_n $$ and $r_i\ge0$ for $i=0,1,\dots,n$, then $r\ge r_0+r_2$.

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It can be done more easily. Set $(\sqrt{n})^{1/n}=1+a_n$, so $$ \sqrt{n}=(1+a_n)^{n}\ge1+na_n $$ by Bernoulli’s inequality. Hence $$ a_n\le\frac{\sqrt{n}-1}{n}=\frac{1}{\sqrt{n}}-\frac{1}{n} $$ so $$ (\sqrt{n})^{1/n}=1+a_n\le 1+\frac{1}{\sqrt{n}}-\frac{1}{n} $$ and the squeeze theorem allows to conclude.

Set $l =\lim_{n\to\infty}n^\frac{1}{n}$

$$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=\sqrt{\lim_{n\to\infty}n^\frac{1}{n}}=\sqrt{l}$$

For the limit of the sequence $n^\frac{1}{n}$ get the sub-sequence $(2n)^\frac{1}{2n}$

So:

$l^2=(\lim (2n)^\frac{1}{2n})^2=\lim (2n)^\frac{1}{n}=\lim 2^\frac{1}{n}. n^\frac{1}{n}=\lim 2^\frac{1}{n}.\lim n^\frac{1}{n}=l$

$l^2=l$ and as $n^\frac{1}{n}\gt1$ we have $l=1$

And $\sqrt{\lim_{n\to\infty}n^\frac{1}{n}}=\sqrt{1}=1$