Non-trivial prime gaps

Question

A simple proof that there are prime gaps of size at least $n+1$ for every $n$ can be seen in the first answer to this question. I consider prime gaps of the form $n!, n!+1, \ldots, n!+n$ of length at least $n+1$ to be "trivial".

My question is:

Can it be proven that for every $n,$ there are prime gaps of length at least $n$ starting from integer $q,$ i.e. $q, q+1, q+2,\ldots, q+n-1,$ such that $q$ is not of the form $n! m,$ where $m$ is positive integer?

I'm not sure if the second answer to the linked question helps answer my question.

Answer 1

A generalisation of the argument might show that there are non-trivial prime gaps.

Suppose that $a_1, \ldots, a_n$ are pairwise relatively prime, and none of them are prime. Then, by the Chinese remainder theorem the system $$x \equiv -1 \ (\text{mod} \ a_1) \quad \cdots \quad x \equiv -n \ (\text{mod} \ a_n)$$ has a solution. Notice that this means that the sequence $x + 1,\ldots, x+n$ are all composite, since $a_i | x+i$. It is left to show that this procedure can yield an $x$ which is not of the form $n! m$, that is $\exists k \leq n$ such that $k$ does not divide $x$. To achieve this, just choose $a_1$ to be a number $\leq n$.

Answer 2

There are no primes with prime gaps being $7$. If you mean the prime gap to be at least $n$, the answer is yes. Consider $n \geq 2$. Take the $n$ smallest distinct primes $p_1, p_2 \dots p_n$. Notice that $q = p_1p_2p_3 \dots p_n + 2$ is not divisible by $n!$ as $3 \nmid q$. However, notice that $\pi (n-1) = k < n$ and thus, we can claim that $q, q+1 \dots q + n - 1$ are all divisible by at least one of $p_1, p_2 \dots p_k$.