TITLE: Conjecture about primes and the factorial: for all primes $p>5$, must there exist a prime $q<p$ such that $q\equiv m!\pmod p$ for some $2<m<p$?
QUESTION [9 upvotes]: Below $0\notin\mathbb N$.
Further corrected conjecture:

For all prime numbers $p>5$ there exist a prime number $q<p$ such that 
  $q\equiv m!\!\pmod p$, $2<m<p$.

or

Given a prime $p>5$ there exist a prime $q<p$ and $k,m\in\mathbb N$, $2<m<p$, such that $kp+q=m!$

I want help to prove the conjecture (which is tested for all $p<100,000,000$) or to find a counter-example.
p=7  q=3  k=3   m=4
p=11 q=2  k=2   m=4
p=13 q=11 k=1   m=4
p=17 q=7  k=1   m=4
p=19 q=5  k=1   m=4
p=23 q=5  k=5   m=5
p=29 q=23 k=173 m=7
p=31 q=7  k=23  m=6
p=37 q=17 k=19  m=6
p=41 q=23 k=17  m=6
p=43 q=29 k=937 m=8
p=47 q=11 k=107 m=7
p=53 q=31 k=13  m=6
p=59 q=2  k=2   m=5
p=61 q=59 k=1   m=5
p=67 q=53 k=1   m=5 ...

Up to the prime 1020361

For several reasons I have had major problems extracting my real observations.  
Also, the primes are not unique having this property, a lot of semiprimes, but not all, and occasionally some other numbers, also have it. It seems like less  than one third of all natural numbers have it and there is a secondary conjecture:

For all primes $p$ there is a prime $q<p^2$ such that $q\equiv m!\pmod {p^2}$,  $2<m<p^2$.

tested for all $p<100,000,000$.

There are strong reasons to believe that the conjecture is true. Suppose 
$0\equiv m!\!\pmod n$, then $0\equiv r!\!\pmod n$ for all $r>m$. And suppose 
$n=sp^t$, where $p$ is the largest prime dividing $n$, $p\nmid s$ and $t>0$, then $0\equiv (pt)!\!\pmod n$. If $p$ is a large prime there are a lot of nonzero solutions to $x\equiv m!\!\pmod p$ and the probability for one of those solutions to be a prime increase with p.

REPLY [6 votes]: For $p=3$, $q$ has to be $2$.
Suppose that there exist $k,m\in\mathbb Z$ such that
$$3k+2=m!$$
Since $m\gt 2$, the RHS is divisible by $3$. This is a contradiction.
Added : Similarly, for $p=5$, there is no such prime $q$.
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