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?
 A: 
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.

Why not? We  can restate Stirling's formula as follows:

There is a function $\psi:(-1,\infty)\to\Bbb R$ (or $\Bbb N_0\to\Bbb R$ if you don't want to worry about the Gamma function) such that: $$x!=x^xe^{-x}\sqrt{2\pi x}\cdot(1+\psi(x))$$For all $x>-1$, and: $$\lim_{x\to\infty}\psi(x)=0$$

So, for any $n\in\Bbb N_0$ and integer $0\le k=k(n)\le n$ we have: $$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n^ne^{-n}\sqrt{2\pi n}(1+\psi(n))}{k(n)^{k(n)}e^{-k(n)}\sqrt{2\pi k(n)}\cdot(1+\psi(k(n))+(n-k(n))^{n-k(n)}e^{k(n)-n}\sqrt{2\pi(n-k(n))}\cdot(1+\psi(n-k(n))}$$
Since $k\in o(n^{2/3})$, it is in particular in $o(n)$ so $k(n)\ll n$ for all large enough $n$, at which point the above expression is valid. Since you are deriving an asymptotic, you don't care about the small $n$, so you may take this as granted.
We can clean this up by diving through by the rightmost factor on the denominator: $$\binom{n}{k}=\frac{\left(\frac{n}{n-k}\right)^{n+1/2}(n-k)^{k}e^{-k}\cdot\frac{1+\psi(n)}{1+\psi(n-k)}}{1+\left(\frac{k}{n-k}\right)^{k+1/2}(n-k)^{-n}e^{n-2k}\cdot\frac{1+\psi(k)}{1+\psi(n-k)}}$$
And we can follow the hint, by taking logarithms. The logarithm of the numerator is: $$-\left(n-k+\frac{1}{2}\right)\log\left(1-\frac{k}{n}\right)+k\log(n)-k+\log\left(\frac{1+\psi(n)}{1+\psi(n-k)}\right)$$Note that $n-k\to\infty$, as $n-k\sim n\to\infty$. The rightmost logarithm then tends to $\log(1)=0$, so I will wrap this up with $o(1)$ (I introduced an 'explicit' function $\psi$ only to emphasise that using asymptotic notation is genuinely rigorous). $-\log(1-k/n)=k/n+\Theta(k^2/n^2)=k/n+o(n^{-2/3})$ (valid, as $k/n\to0$). We get: $$\frac{k-k^2}{n}+\frac{1}{2}\varphi(n)+k\log(n)-k+o(1)$$Using $n-k\sim n$ again, where $\varphi(n)\sim k^2/n$.
How does the denominator behave? You need to check the denominator tends to $1$ and use $\log(1+\cdots)$'s expansion again. It will probably be quite tedious, but this approach should give you the right answer if you do it carefully (see, I could have ignored $\varphi(n)$ and written $\Theta(k^2/n)$, but I see the target expression features $k^2/2n$ so it seems as if more detail is needed!).
A: 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 $\sqrt{2\pi n}$.
A: This does not answer exactly the question
Consider the case where $k=n^a$ with $0\lt a \lt 1$
$$\binom{n}{k}=\binom{n}{n^a}=\frac{\Gamma (n+1)}{\Gamma \left(n+1-n^a\right)\, \Gamma  \left(n^a+1\right)}$$
Taking logarithms and using three times Stirling approximation, it is simple to obtain
$$\log \Bigg[\binom{n}{n^a}\Bigg]=\big[1+(1-a)\log(n)\big]n^a-\frac a2 \log(n)-\frac{1}{2} \log (2 \pi )+O\left(\frac{1}{n^a}\right)$$
