I was recently studying about reduced residue systems , and then I stumbled upon an idea.
Let $P$ represent the set of all primes.
Let $P_{n}\#$ denote the product of the first $n$ primes and let $P_{n}$ denote the $n^{th}$ prime.
Let $S$ be a set such that, $S=\{ (x_{1},x_{2}) : x_{1},x_{2} \in P \wedge P_{n-1}^{2} < x_{1}\cdot x_{2} < P_{n}\# \}$
My conjecture is:
$P_{n}\# - 1$ is a prime if,
$P_{n}\# \not \equiv K\cdot A + 1 \pmod{P_{n-1}\#} $ $\forall$ $(K,A) \in S$
For example: Let's take $n=3$.
$P_2\# =6$
$S=\phi$
Hence, $P_3\# -1 =29$ is prime.
Let's take $n=4$.
$P_3\# =30$
$S=\{(7,7),(11,7),(7,11),(11,11),(11,13),(13,11),(13,13),(7,17),(17,7),(11,17),(17,11),(7,19),(19,7),(11,19),(19,11)\}$
Hence, $P_4\# -1 =209$ ($209 = 19 \cdot 11$) is not prime since $(7,17)$, $(17,7)$, $(19,11)$, $(11,19)$ satisfy the conditions.
Similarly, $P_{7} \# -1$ is not prime since $P_{7} \# \equiv 41\cdot 11719$ (mod $P_{6} \#$)
If there are any proofs or counter examples for this conjecture, it would be of great help.