Denote $p_n$ be the $n$th prime number.
Theorem(Posa) For any$k\in\mathbb{N}$,there is a constant $N$ such that each $n>N$,have $$p_1p_2\cdots p_n>p_{n+1}^k\quad (*)$$
My question:How can we find the largest nature number $n$,such that $$p_{n+1}^k>p_1\cdots p_n$$ More then ,is there a exact formula about the least constant $N$ such that $(*)$holds？ I think we must find a more exact bound of $p_n$.
For example：when $k=3$ ，note Bertrand's postulate gives
$$p_{n+1} \lt 2p_{n} \lt 4p_{n-1} \lt 8p_{n-2} $$
Thus,
$$p_{n+1}^3 \lt (2p_{n})(4p_{n-1})(8p_{n-2}) = 64(p_{n-2}p_{n-1}p_{n}) $$
For $n = 5$,
$$13^3 = 2179 \lt 2(3)(5)(7)(11) = 2310 $$
so it's true. For $n = 6$,
$$17^3 = 4913 \lt 2(3)(5)(7)(11)(13) = 30030 \tag{5}\label{eq5A}$$
so that is also true. For $n \ge 7$,
$$\prod_{i=1}^{n}p_i \ge (p_1 p_2 p_3 p_4)p_{n-2}p_{n-1}p_{n} = 210(p_{n-2}p_{n-1}p_{n})$$
Since $210 \gt 64$, is also true for all $n \ge 7$.So for $n\ge 5$， $$p_1p_2\cdots p_n>p_{n+1}^3$$ and the least constant$N=5$.
Its equivalent to $\ln p_1 p_2 \cdots p_n > k \ln p_{n+1}$ and so pick $ k < \frac{\theta(p_n)}{\ln p_{n+1}} \approx \frac{p_n}{\ln p_n} $