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 ...
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.