Fix a positive integer $k$. Let $$ f(n):= \frac{k!\binom{n}{k}}{n^k} $$ Then $\lim\_{n\to \infty} f(n) = 1$. Hence $f(n) \ge 1-\epsilon$ for large $n$.


Define $n\_0(\epsilon)$ as the least positive integer such that $n\ge n\_0$ implies $f(n)\ge 1 -\epsilon$. Is there an asymptotic for $n\_0(\epsilon)$ as $\epsilon\to 0$? A lower bound for $n\_0(\epsilon)$ for fixed small $\epsilon$ would be useful too.



Is there an asymptotic formula for an inverse function of the binomial coefficient?
The approximation coming from the normal approximation of the binomial distribution is
$$ k = \sqrt{-2n\ln(1-\epsilon)}.$$
To get more terms write $\Delta=-\ln(1-\epsilon)$, then by expanding $-\ln(1-i/n)$ as a Taylor series in $i$ and summing over $i=0\ldots k-1$, you get a series 
$$ \Delta = 
1/2\,{\frac {k\, ( k-1 ) }{n}}+1/12\,{\frac {k\, ( 2\,
k-1) ( k-1 ) }{n^2}}+1/12\,{\frac {{k}^{2}
 ( k-1) ^{2}}{{n}^{3}}} + \cdots
$$
which converges if $k=O(n^c)$ for $c<1$ (but is useless if $c$ is too close to 1). The dominant term for moderately small $k$ is $k^2/2n$. You can use a series reversion method to get more terms of the inverse. I think (without much checking) that
$$ k = \sqrt{2n\Delta} + \frac{3-2\Delta}{6} + \text{smaller terms},$$
and note that we are already running into the issue of $k$ being an integer.


