It's well-known that for each prime number $p$ there are exactly two groups of order $p^2$, five of order $p^3$, and fifteen of order $p^4$ (at least when $p>3$).
I know that the classification of $p$-groups gets much harder for higher exponents. But I wonder if $p$-groups with higher dimensions are still structured enough that they would at least theoretically allow a classification.
In particular: Is there, for every natural $n$, a prime $p_0(n)$ and a number $m(n)$ such that for every $p \ge p_0(n)$ the number of groups of order $p^n$ equals $m(n)$?
(For example, $m(2)=2,m(3)=5,m(4)=15$ for $p_0(2)=p_0(3)=2$ $p_0(4)=5$.)