Consider the product $$\prod_p\Big(1+\frac1{(p-1)^3}\Big)$$ over the primes. I can't easily see why this should converge. Something like $\prod_p\big(1+\frac 1{p-1}\big)$ doesn't converge, since (I believe) $\prod_{p\leqslant x}\big(1+\frac 1{p-1}\big)\sim\log\log x$.
I've rewritten the product as $$\exp\bigg(\sum_{p}\sum_{n}\frac{(-1)^{n+1}}{n(p-1)^{3n}}\bigg)$$ using the Taylor series for log, but I'm not sure how to continue.
On
Since, for $x > 0$, $\ln(1+x) =\int_0^x \dfrac{dt}{1+t} $, $\ln(1+x) \lt x$ and $\ln(1+x) \gt \dfrac{x}{1+x} \gt x-x^2 $.
Therefore
$\begin{array}\\ \prod_p\Big(1+\frac1{(p-1)^3}\Big) &=e^{\sum_p \ln\Big(1+\frac1{(p-1)^3}\Big)}\\ &\lt e^{\sum_p \frac1{(p-1)^3}}\\ \end{array} $
and since $\sum_p \frac1{(p-1)^3}$ converges, $e^{\sum_p \frac1{(p-1)^3}} $ converges.
More generally, if $a_n > 0$ and $\sum a_n$ converges then
$\begin{array}\\ \prod_n\Big(1+a_n\Big) &=e^{\sum_n \ln(1+a_n)}\\ &\lt e^{\sum_n a_n}\\ \end{array} $
so $\prod_n\Big(1+a_n\Big)$ converges.
Let $p$ denote the $n$th prime so$$p-1\ge n\implies1+(p-1)^{-3}\le\exp n^{-3},$$so your product $\le\exp\zeta(3)$.