Given that: $$n={k^2 \choose k}$$ what is $k$ as a function of $n$?
So far, I have found the following approximation:
$$ n \approx (k^2)^k = (k^k)^2 $$ $$ \sqrt{n} \approx k^k $$
If we take this approximation as an equality, then according to this answer:
$$ k = {\ln{\sqrt{n}} \over W(\ln{\sqrt{n}})} $$ where $W$ is Lambert's function.
Asymptotically, $W(z) = \Theta(\ln z)$, so we get:
$$ k = \Theta\left({\ln n \over \ln \ln n}\right) $$
Is this approximation correct (asymptotically)? Is there a better approximation?