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 $n$.

Is there an asymptotic for $n_0(\epsilon)$, defined as the least positive integer such that $n\ge n_0$ implies $f(n)\ge 1 -\epsilon$?

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.