A conjecture about primorials A primorial, denoted $p_n\#$, is the product of the first $n$ prime numbers ($p_1=2,\ p_2=3$ etc.). The magnitude of primorials grows rapidly beyond the range of convenient arithmetic manipulation.  The number $(p_n\#+1)$ is not divisible by any of the first $n$ primes, and so is frequently a prime number itself. For $n=1,2,3,4,5,11$, $(p_n\#+1) \in \mathbb P$.
I noticed (for primorials accessible to calculation) that when $(p_n\#+1) \not \in \mathbb P$, that a 'near primorial' number plus $1$ could be identified that was a prime. By near primorial number, I mean the product of all but one of the first $n$ primes, or $\frac{p_n\#}{p_i};\ 1<i<n$. For example, $\frac{p_8\#}{3}+1,\ \frac{p_{10}\#}{3}+1,\ \frac{p_6\#}{5}+1,\ \frac{p_7\#}{5}+1,\ \frac{p_{12}\#}{7}+1,\ \frac{p_{13}\#}{11}+1,\ \frac{p_{9}\#}{13}+1$ are all primes. Examples of this kind can be rewritten in the form $p_n\#=p_i(p_k-1);\ 1<i<n,\ k>n$.
Based on this admittedly extremely small set, I conjecture that it might be the case $$p_n\#=C(p_k-1);\ C\in \{1,p_i\},\ 1<i<n,\ k>n$$ The signal feature of $C$ is that it is not composite. A single counterexample arrived at by computation would disprove the conjecture, but for $p_{14}\#$ and greater, the numbers are beyond my ability to conveniently calculate.
My questions are: Has this conjecture been previously considered and settled? If not, is there an analytic approach to prove or disprove the conjecture? 
 A: What says the random model : 
Let $j\le n$ and $$f(j,n) = 1+\prod_{i=1, i \ne j}^n p_i$$
By Mertens theorem $\log f(j,n) \approx \sum_{i=1}^n \log p_i \approx n$
Assuming independence of the congruences $\bmod$ different primes
$$Pr(f(j,n) \text{ is prime}] \approx \frac{\prod_{i \ne j} (1-p_i)^{-1}}{\ln N} \ge C\exp(\sum_{i=1}^{n-1}\frac{1}{p_i} - \ln \ln N)\\ \approx C \exp( \ln \ln (n-1) - \ln n) \approx C\frac{\ln n}{n}$$
Taking $j $ uniformly in $1\ldots n$, assuming the random variables "$f(j,n)$ is prime" are independent,
the probability that none of the $f(j,n)$ is prime is $$\approx \prod_{j=1}^n (1-C\frac{\ln n}{n})= (1-C\frac{\ln n}{n})^n = \exp(n\log (1-C\frac{\ln n}{n})) \approx \exp(-C \ln n)) = n^{-C}$$
If you redo it replacing $j$ by a subset $J \subset 1 \ldots n$ with $4$ elements and $f(j,n)$ by $f(J,n) = 1+\prod_{i=1, i \not \in J} p_i$ you'll get $C > 1$ so that the probability that for some $n \ge N$, none of the $f(J,n)$ is prime is $\le \sum_{n=N}^\infty n^{-C}$ which $\to 0$ as $N \to \infty$,
ie. it is almost surely true that for every $n$ large enough $p_n\# = a_n (p_{k_n}-1)$ with $a_n$ a product of at most $4$ primes.
