The falling (descending) factorial $x^\underline n$ for $n,x\in\mathbb N, x>0$ is defined as:
$$x^\underline n = \prod_{k=0}^{n-1}(x-k).$$
What is known about the specialization of the falling factorial in which the summands are the primes, i.e.,
$$ x^\underline {\overline{n}} = \prod_{k=1}^{n-1}(x-p_k), $$
where $p_k$ is the $k ^{\rm th}$ prime?
Generalizations of the rising/falling factorial have been previously studied; see e.g. this article or this Wikipedia section.
Quoting some of the results from those articles, we have: $\newcommand{\stirling}{\genfrac{[}{]}{0pt}{}}$ Let $\stirling{n}{k}$ denote the coefficient of $x^{k-1}$ in the $n$th falling prime factorial. Then the coefficients satisfy the following relation for $n\geq 1$: $$\stirling{n}{k}=\stirling{n-1}{k-1}-p_{n-1}\stirling{n-1}{k}$$
We also have interesting identities on the prime harmonic numbers such as: $$\left(\prod_{j=1}^n p_j^2\right)\sum_{k=1}^n\frac{1}{p_k^2}=\stirling{n+1}{2}^2-2\stirling{n+1}{1}\stirling{n+1}{3}$$
See the references for a lot of relations along those lines.