Follow-up regarding disproof of modified conjecture found in MathOverflow

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I was looking through the OEIS, and I found the following conjecture on sequence $\text{A}007097$:

If $a_n$ is a sequence for which $a_1=2$ and $a_{n+1}$ is the $a_n$-th prime, $$\prod_{i=1}^{n}\log(a_i)\sim a_n$$

I then saw that this conjecture had been disproved on MathOverflow by GH from MO in the comments on this question, but with a substituted possible replacement conjecture. The solver claimed the replacement was probably false as well, but that a proof would be considerably more difficult. So my question is, is the replacement conjecture true? Or more specifically, determine the veracity of the following:

If $a_n$ is a sequence for which $a_1=2$ and $a_{n+1}$ is the $a_n$-th prime, $$\prod_{i=1}^{n}\log(a_i)\sim a_{n+1}$$

My own math skills are decidedly insufficient to determine this question one way or the other, but I'm really interested in the problem and would be curious to see a resolution. Any help would be appreciated. Thanks!

Edit: I've learned that the person who "disproved" the original conjecture on MathOverflow has since then realized that they made a mistake in their disproof, so the original conjecture may still be valid. Maybe a fun challenge for someone else.

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Let, for each $n \geq 1$, $p_n$ be the $n$-th prime number, and $1+r_n=\frac{p_n}{n\ln{n}}$, so that $r_n \sim \frac{\ln{\ln{n}}}{\ln{n}}$ by refinements of the prime number theorem.

Let $\delta_n=r_{a_n}$, so that $\ln{a_n}=\frac{a_{n+1}}{a_n(1+\delta_n)}$.

Taking products, it follows $a_{n+1}=2\prod_{k=1}^n{(1+\delta_k)}\prod_{k=1}^n{\ln{a_k}}$.

Now we’ll show that $\prod_{k=1}^n{(1+\delta_k)}$ diverges. It’s enough to show that $\sum_{k \geq 1}{\frac{\ln{\ln{a_k}}}{\ln{a_k}}}$ diverges.

Let $b_k=\ln{a_k}$, clearly $b_k$ grows to infinity and by the prime number theorem $b_{k+1}-b_k=\ln{b_k}+o(1)$. We want to show that $\sum_{k \geq 1}{\frac{\ln{b_k}}{b_k}}$ diverges to infinity. But as $\frac{\ln{b_k}}{b_k}\sim \frac{b_{k+1}-b_k}{b_k} \sim \ln{\frac{b_{k+1}}{b_k}}$, this is clearly divergent.

By doing everything explicitly, we can even write $a_{n+1}=2\left(\prod_{k=1}^n{\ln{a_k}}\right)(b_{n+1}/b_1)^{1+o(1)}$.