I do not understand the very last equality before the arrow. It seems to pop up out of nowhere. Below I copy the proof in its entirety. This is from Buck's Advanced Calculus text pg. 46. Perhaps, this is a silly misunderstanding, but to me it doesn't seem to follow from the rest of the proof.
$ \it \bf Lemma: $
$$ \lim_{n \to \infty} (\frac{ \log n}{n}) = 0$$
$ \it \bf Proof: $ Given $n$, choose $ k $ so that $(k-1)^2 \leq n < k^2$. It is easily checked that the inequality $k^2 < 2^{k-1}$ holds for all $k \geq 7$. Thus, for any $n \geq (6)^2 = 36$, we have
$$ \log n \leq \log(k^2) < \log(2^{k-1}) = (k-1) \log 2 \leq \sqrt{n} \log 2 $$ and we have shown that $$ \frac{ \log n}{n} \leq \frac{ \sqrt{n}\log 2}{n} = \frac{ \log 2}{\sqrt{n}} \rightarrow 0$$.
Since for $n\in \mathbb{R}^+$ you can write $n = \sqrt{n}\sqrt{n}$, then
$$\frac{\sqrt{n}\log(2)}{n} = \frac{\sqrt{n}\log(2)}{\sqrt{n}\sqrt{n}} = \frac{\log(2)}{\sqrt{n}}$$