How to prove an asymptotic estimate for $n \choose k$, where $k = \mathcal{o}(n^{2/3})$? 
Let $k = k(n)$ such that $k \rightarrow \infty$ as $n \rightarrow \infty$, but $k = \mathcal{o}(n^{2/3})$. Use Stirling's formula $$n! = (1+\mathcal{o}(1))\sqrt{2\pi n} \biggl(\frac{n}{e}\biggr)^n$$
and the Taylor expansion of $\ln(1+x)$ to show that
$${n\choose k} = (1+\mathcal{o}(1))\frac{1}{\sqrt{2\pi n}} \biggl(\frac{n}{k}\biggr)^k \begin{cases} \exp(k) \quad \text{ for } k = \mathcal{o}(n^{1/2}) \\ \exp \biggl(k-\frac{k^2}{2n}\biggr) \quad \text{ for } k = \Omega(n^{1/2})\end{cases} .$$

I understand that by definition we have
$${n\choose k} = \frac{n!}{k! (n-k)!},$$
but I guess we can not just simply use Stirling's formula on $n!$, $k!$ and $(n-k)!$ at the same time, so I do not see how to proceed. Could you please give me a hint?

It follows from the main result of this paper that
$$
\log ((n + a)!) = \left( {n + a + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \frac{{6a^2  + 6a + 1}}{{12n}} + \mathcal{O}\!\left( {\frac{{\max (\left| a \right|^3 ,1)}}{{n^2 }}} \right)
$$
provided $n + a+1 \ge 0$ and $\left| a+1 \right| < \frac{3}{5}n$, where the implied constant does not depend on $n$ or $a$. Then
$$
\log (n!) = \left( {n + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + o(1),
$$
$$
\log (k!) = \left( {k + \frac{1}{2}} \right)\log k - k + \frac{1}{2}\log (2\pi ) + o(1),
$$
$$
\log ((n - k)!) = \left( {n - k + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log (2\pi ) + \frac{k^2}{{2n}} + o(1),
$$ as $n,k\to +\infty$ with $k=o(n^{2/3})$. Accordingly,
$$
\log \binom{n}{k} =  - \frac{1}{2}\log (2\pi k) + k\log \left( {\frac{n}{k}} \right) + k - \frac{k^2}{{2n}} + o(1)
$$
or
$$
\binom{n}{k} = \frac{1}{{\sqrt {2\pi k} }}\left( {\frac{n}{k}} \right)^k \exp\! \left( {k - \frac{k^2}{{2n}}} \right)(1 + o(1)),
$$
as $n,k\to +\infty$ with $k=o(n^{2/3})$. Note that the factor involves $\sqrt{2\pi k}$ and not $\sqrt{2\pi n}$.
