A conjecture about primorials

A few lines of Mathematica shows that $p_{19}\#$ is the first counterexample. $$p_{19}\# = \bigg(\prod_{i=1}^{19}p_i\bigg)+1=7858321551080267055879091=54730729297\cdot 143581524529603,$$ so it is composite. The following table shows that $\frac{p_{19}\#}{p_n}+1$ is composite for all $n$ satisfying $1\leq n < 19$
$$\begin{array}{|c|c|c|c|} n & p_n & \frac{p_{19}\#}{p_n}+1 & \text{smallest divisor of }\frac{p_{19}\#}{p_n}+1\\ \hline 1 & 2 & 3929160775540133527939546 & 2 \\ \hline 2 & 3 & 2619440517026755685293031 & 613 \\ \hline 3 & 5 & 1571664310216053411175819 & 5501 \\ \hline 4 & 7 & 1122617364440038150839871 & 21713 \\ \hline 5 & 11 & 714392868280024277807191 & 389 \\ \hline 6 & 13 & 604486273160020542759931 & 131 \\ \hline 7 & 17 & 462254208887074532698771 & 101 \\ \hline 8 & 19 & 413595871109487739783111 & 136483 \\ \hline 9 & 23 & 341666154394794219820831 & 26801 \\ \hline 10 & 29 & 270976605209664381237211 & 809 \\ \hline 11 & 31 & 253494243583234421157391 & 127 \\ \hline 12 & 37 & 212387068948115325834571 & 3449 \\ \hline 13 & 41 & 191666379294640659899491 & 3593 \\ \hline 14 & 43 & 182751663978610861764631 & 167 \\ \hline 15 & 47 & 167198330874048235231471 & 71 \\ \hline 16 & 53 & 148270217944910699167531 & 2866463 \\ \hline 17 & 59 & 133191890696275712811511 & 283 \\ \hline 18 & 61 & 128824943460332246817691 & 179 \\ \hline \end{array}$$

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.