With $p_i$ being the $i$-th prime, I'm wondering whether there is a tighter bound than $\alpha = 4$ in the relation $$ \prod_{i=1}^n p_i < \alpha^{p_n} $$ $\alpha = 4$, which is tight enough to be used to prove Bertrand's postulate (Chebyshev's theorem) that there is always a prime in the interval $[k,2k]$, is easily proven using $\binom{2r}{r} < 4^r$ and the fact that any prime in the interval $[n/2,n]$ will divide $\binom{2r}{r} < 4^r$; see the answer to

https://math.stackexchange.com/questions/1924453/show-that-p-1p-2-cdots-p-t4n

However, that argument makes a couple of lavishly conservative statements, two of which are that since $p_n$ divides $\binom{2n}{n}$ it must be less than $\binom{2n}{n}$, and that $\binom{2n}{n} < \sum_k \binom{2n}{k}$. One might suspect that a smaller value of $\alpha$ could suffice.

Numerical experimentation finds that for $n \leq 100$,

$$ \prod_{i=1}^n p_i < (2.62)^{p_n} $$ but if you extend to $n=1000$, you need to increase to $\alpha = 2.6938$.

Thus my question: Is it known that no $\alpha < 4$ works in the inequality as $n\to \infty$, has there been a lower value demonstrated, or is this question completely open?

I would guess that this can be attacked using the Prime Number Theorem, but fluctuations from that distribution become important for this bound.