Find asymptotics in a given form $n=(e+o(1))^{f(s)}$ 
Let $p\to\infty$, $s={\binom {p^4} p}$ and $n={\binom {p^4}{p^2}}$. Find a function $f(s)$ in the following form $$\large n=(e+o(1))^{f(s)}$$

I've tried to use the followinf asymptotics for binomial coeffs:
$$\binom n k =  \frac{n^k \exp\left(k^2/2n\right)}{k!}(1+O(1))$$
And I've got the following results:

$$s={\binom {p^4} p} = \exp\left({2p^2 \ln p - \frac{1}{2} (1 + \ln 2\pi) + p^2 + \ln p + O\left(\frac{1}{p^2}\right)}\right)$$
$$n={\binom {p^4}{p^2}} = \exp\left({4p\ln p - \frac{1}{2p^2} - \left(p + \frac{1}{2}\right)\ln p + p - \frac{1}{2}\ln 2\pi + O\left(\frac{1}{p}\right)}\right)$$

What should I do next? A hint would be useful!
 A: For the beginning some preparatory computations:
$$
\begin{align}
{n\choose k}
&=\frac{n\ldots (n-k+1)}{k!}\\
&=\frac{n^k}{k!}\left(1-\frac{1}{n}\right)\ldots\left(1-\frac{k-1}{n}\right)\\
&=\frac{n^k}{k!}\exp\left(\sum_{i=1}^{k-1}\ln\left(1-\frac{i}{n}\right)\right)\\
&=\frac{n^k}{k!}\exp\left(\sum_{i=1}^{k-1}\left(-\frac{i}{n}+O\left(\frac{i^2}{n^2}\right)\right)\right)\\
&=\frac{n^k}{k!}\exp\left(-\frac{k(k-1)}{2n}+O\left(\frac{k^3}{n^2}\right)\right)\\
\end{align}
$$
Thus
$$
{n\choose k}=\frac{n^k}{k!}e^{-\frac{k(k-1)}{2n}+O\left(\frac{k^3}{n^2}\right)}
$$
Given this formula and Stirling's approximation we get
$$
\begin{align}
\ln s
&=\ln {p^4\choose p}\\
&=p\ln p^4-\ln p!-\frac{p(p-1)}{2p^4}+O\left(\frac{p^3}{p^8}\right)\\
&=4p\ln p-\left(p\ln p-p+\frac{1}{2}\ln p+\ln\sqrt{2\pi}+O\left(\frac{1}{p}\right)\right)-\frac{1}{2p^2}+\frac{1}{2p^3}+O\left(\frac{1}{p^5}\right)\\
&=3p\ln p+p-\frac{1}{2}\ln p-\frac{1}{2}\ln 2\pi+O\left(p^{-1}\right)\\
&=3p\ln p+O(p)
\end{align}
$$
$$
\begin{align}
\ln n
&=\ln {p^4\choose p^2}\\
&=p^2\ln p^4-\ln (p^2)!-\frac{p^2(p^2-1)}{2p^4}+O\left(\frac{p^6}{p^8}\right)\\
&=4p^2\ln p-\left(p^2\ln p^2-p^2+\frac{1}{2}\ln p^2+\ln\sqrt{2\pi}+O\left(\frac{1}{p^2}\right)\right)-\frac{1}{2}+\frac{1}{2p^2}+O\left(\frac{1}{p^2}\right)\\
&=2p^2\ln p+p^2-\ln p-\frac{1}{2}\ln 2\pi e+\frac{1}{2p^2}+O\left(p^{-2}\right)\\
&=2p^2\ln p+O(p^2)
\end{align}
$$
We claim, that
$$
f(x)=\frac{2\ln^2 x}{9\ln\ln x}
$$ 
is the desired function. Indeed,
$$
\begin{align}
\ln^2 s
&=(3p\ln p+O(p))^2\\
&=9p^2\ln^2 p+6p\ln p O(p)+O(p)^2\\
&=9p^2\ln^2 p+ O(p^2\ln p)+O(p^2)\\
&=9p^2\ln^2 p+O(p^2\ln p)\\
\end{align}
$$
$$
\begin{align}
\ln\ln s
&=\ln(3p\ln p+O(p))\\
&=\ln(3p\ln p(1+O(\ln^{-1}p)))\\
&=\ln 3+\ln p + \ln\ln p + \ln(1+O(\ln^{-1}p))\\
&=\ln p + \ln\ln p+\ln 3+O(\ln^{-1} p)\\
&=\ln p+O(\ln\ln p)\\
\end{align}
$$
$$
\begin{align}
f(s)
&=\frac{2\ln^2 s}{9\ln\ln s}\\
&=\frac{2(9p^2\ln^2p+O(p^2\ln p))}{9\ln p+O(\ln\ln p)}\\
&=\frac{9\ln p(2p^2\ln p+O(p^2))}{9\ln p\left(1+O\left(\frac{\ln\ln p}{\ln p}\right)\right)}\\
&=(2p^2\ln p+O(p^2))\left(1+O\left(\frac{\ln\ln p}{\ln p}\right)\right)\\
&=2p^2\ln p+O(2p^2\ln\ln p)+O(p^2)+O\left(p^2\frac{\ln\ln p}{\ln p}\right)\\
&=2p^2\ln p+O(p^2)\\
\end{align}
$$
Since
$$
\ln n-f(s)=2p^2\ln p+O(p^2)-(2p^2\ln p+O(p^2))=O(p^2)=o(2p^2\ln p+O(p^2))=o(f(s))
$$
then $n=(e+o(1))^{f(s)}$
