Just playing with the result in this answer, I am asking some help for computing:
$$\lim_{n \to \infty}\frac{\prod_{k=1}^{n} p_{k}^{\frac{p_k}{(p_k-1)^2}}}{p_n}$$
where $p_k$ is the $k$-th prime number. The sequence computed at $n=16000$ evaluates to about $1.9180508$.
The aim of this answer is to obtain an analytic expression for the limit asked by OP. By a change of notation, we have
$$ L=\lim_{n\to\infty}{\prod_{k=1}^np_k^{p_k\over(p_k-1)^2}\over p_n}=\lim_{x\to\infty}L(x), $$
where
$$ L(x)=x^{-1}\prod_{p\le x}p^{p\over(p-1)^2}.\tag1 $$
Taking logarithms gives
$$ \log L(x)=\sum_{p\le x}{p\over (p-1)^2}\log p-\log x.\tag2 $$
By the prime number theorem, we have
$$ \sum_{p\le x}{\log p\over p}=\log x-C+O\left(1\over\log x\right),\tag3 $$
in which
$$ C=\gamma+\sum_p{\log p\over p(p-1)}. $$
Plugging (3) into (2), we have
\begin{aligned} \log L(x)&=\sum_{p\le x}{\log p\over p}-\log x+\sum_{p\le x}\left[{p\over(p-1)^2}-\frac1p\right]\log p \\ &=-C+\sum_{p\le x}{2p-1\over p(p-1)^2}\log p. \end{aligned}
Since the term in the last sum is $\ll(\log p)/p^2$, it converges as $x\to+\infty$. Plugging this into (1) gives
$$ \log L=-\gamma+\sum_p{\log p\over(p-1)^2}.\tag4 $$
Numerical computations
Let sequence $L_N$ be defined by
$$ \log L_N=-\gamma+\sum_{p\le N}{\log p\over(p-1)^2}. $$
Then we have
\begin{aligned} R_N &=\log{L\over L_N}=\sum_{p>N}{\log p\over(p-1)^2} \\ &<\int_N^{+\infty}{\log t\over(t-1)^2}\mathrm dt={\log N\over N-1}-\log\left(1-\frac1N\right). \end{aligned}
From Mathematica computation, we have for $N=10^8$ that
$$ 1.9150679<L_N<L<e^{R_N}L_N<1.9150684. $$