2015 Danube Mathematical Competition

Notice that $5$ is the first index $k$ such that $p_1\cdots p_{k-1}>p_k^2$. Now, if $p_1\cdots p_{k-1}>p_k^2$ for some index $k\ge 5$, then $p_1\cdots p_{k-1}{p}_k>p_k^3>4p_k^2>p_{k+1}^2$, by the Bertrand-Tchebysheff theorem, so $p_1\cdots p_{k-1}>p_k^2$ for all indices $k\ge 5$.

Consequently, $m\le 4$, so $n<p_4^2=49$. Examination of the possible cases quickly yields the required numbers: $1,2,3,4,6,8,12,18,24,30$.