Is a factorial-primorial mesh divisible by the factorial infinitely often? Question
Suppose the factorial and primorial functions conceived a baby through the act of addition and called it $n!\#$, and it looked like this:
$$n!\# = \prod_{i=1}^n (p_i + i) = (2 + 1)(3 + 2)(5 + 3)(7 + 4) \dots (p_n + n)$$
Does it happen infinitely often that the factorial divides this function, i.e. are there infinite integer solutions to:
$$\dfrac{n!\#}{n!}$$
Background
The factorial function $n!$ is well known and is given by:
$$\prod_{i=1}^n i = 1 \times 2 \times 3 \times \dots n$$
The marginally less well known primorial function $p_n\#$ is given by:
$$\prod_{i=1}^n p_i = 2 \times 3 \times 5 \times \dots p_n$$
where $p_i$ is the $i^{th}$ prime.
The function defined above is catalogued in the OEIS under the following link: OEIS reference. However there is no reference to divisibility by $n!$ there.
What I know so far
Here are the first few values I've computed for $n!\#$:
$\begin{array}{c|ccccc}
n & n! & n!\# & \textrm{divides?} \\ 
\hline
1 & 1 & 3 & \textrm{Yes} \\ 
2 & 2 & 15 & \textrm{No} \\ 
3 & 6 & 120 & \textrm{Yes}\\ 
4 & 24 & 1320 & \textrm{Yes} \\
5 & 120 & 21120 & \textrm{Yes} \\
6 & 720 & 401280 & \textrm{No} \\
7 & 5040 & 9630720 & \textrm{No} \\
\end{array}$
My Intuition
I'm of two minds as to the truth of this statement, and I don't have enough experience with numbers to judge either way, so I'm on the fence for now. Here are my two basic arguments, which are far from rigorous.
Argument for: The primes are randomly scattered, whereas the sequence $1, 2, 3, \dots, n$ isn't. So adding $p_n + n$ should preserve the randomness that was there in the primes in the first place. By randomness, I mean there should be no preference shown for particular prime factors over others. Now, since $n\#!$ grows much faster than $n!$, it should eventually start sweeping up all the primes in $n!$. Perhaps there is even some $N$ such that that divisibility of $n!\#$ by $n!$ is true for all $n > N$, but this is quite strong and I'm not so sure.
Argument against: On the other hand, what throws doubt on the conjecture is that because $n\#!$ grows so fast, perhaps it grows too fast, and misses lots of little primes that are bundling up in $n!$. In other words, perhaps there is a point $N$ such that that divisibility of $n!\#$ by $n!$ is false for all $n > N$.
 A: This is not a proof, but here is some relevant numerical evidence for the conjecture that $n=1,3,4,5$ are the only times (not counting $n=0$) this fraction is an integer.
Firstly, it's the only solution for $n$ up to $15000$. (All Mathematica code below.)
More importantly, the number of prime factors in the denominator that are not cancelled out (counted with multiplicity so that $20=2^2*5$ has $3$ prime factors) appears to grow roughly linearly with $n$, yet it would have to drop to $0$ for the fraction to be an integer. 
Here is an unconvincing plot for $0\le n<100$:

Here is a plot for $0\le n\le7000$ with the line $y=(0.0868166)x+27$ drawn on top:

(*Mathematica Code*)

f[0]=1;
f[n_]:=f[n]=f[n-1]*(Prime[n]+n);

Print[Table[ If[IntegerQ[f[n]/(n!)], n, Nothing], {n, 0, 15000}]];

countprimes[n_] := countprimes[n] = -Total[Transpose[Select[FactorInteger[f[n]/(n!)], Function[x, 0 > x[[2]] ] ]][[2]]];

Print[ListPlot[Table[{n, If[n==0||n==1||n==3||n==4||n==5,0,countprimes[n]]}, {n, 0, 99}]]];

m=Table[{n, If[n==0||n==1||n==3||n==4||n==5,0,countprimes[n]]}, {n, 0, 7000}];

Print[LinearModelFit[m, x, x]];

Show[ListPlot[m], Plot[27 + 0.0868166 x, {x, 0, 7000}, PlotStyle -> Orange]]

A: Conjecture: for all prime $q$, $q$ divides $p_n + n$ infinitely often.  We know by Dirichlet's theorem that there are an infinite number of primes in any arithmetic progression, $p_n = a \cdot q + b$ with $a$ free and $0 \lt b \lt q$, and there is no apparent relationship between $b$ and $n$.  But this looks like it might be very difficult to prove. 
This is equivalent to: for all $k$, there exists an $n$ such that $\frac{ n!\# }{k!}$ is an integer; a much weaker version of your problem, so proving the affirmative will be at least as hard.
If it is not true that $n!$ divides $n!\#$ infinitely often, we should first be able to prove that infinitely often $n!$ does not divide $n!\#$ and so far I'm unable to do that either.  Heuristically, the $n$ terms of $n!\#$ are random $\text{mod} \space n$, so we should expect a $(1 - \frac{1}{n})^n \rightarrow \frac{1}{e}$ probability that none of them are divisible by $n$, and when $n$ is prime this means at least a $\frac{1}{e}$ chance that $n!$ does not divide $n!\#$.  I think this can be turned into a heuristic argument against your original statement.
I wrote a program to check the condition.  The only values I found where $n! \space \vert \space n!\#$ and $n \lt 3000$ are $n = 1, 3, 4, 5$.  That means the answer is likely "no", to learn more about why it might be interesting to study the resulting denominator at each stage.
I also tried it with primorial instead of factorial in the denominator, I found $n = 1, 3, 4, 5, 6, 11, 12, 13, 14, 15, 16$ and no others less than $3000$.  It's not just the multiplicity of smaller primes in the denominator that's causing divisibility to fail, the introduction of new primes seems to be enough.
I'll update this answer if I make any more progress.
