Danube Mathematical Competition, 2015

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: $1, 2, 3, 4, 6, 8, 12, 18, 24, 30$.