Conjecture concerning modular arithmetic

Question

Below $0\notin\mathbb N$. I want a proof or a counter-example of the following (corrected) conjecture:

Suppose $p$ is the smallest prime dividing $n\in\mathbb N$ and suppose $kn+ap=m!$, where $ap<kn$, for some $a,k,m\in\mathbb N$, then there are $k^{\prime},a^{\prime}\in\mathbb N$ such that $k^{\prime}n+a^{\prime}p=(m+1)!$, where $a^{\prime}p<k^{\prime}n$.

A more simple way to write the conjecture is (with some abuse of language):

Suppose $p$ is the smallest prime dividing $n\in\mathbb N$ and suppose $p|(m!\pmod n)$, then $p|((m+1)!\pmod n)$

Answer 1

Your corrected conjecture is true.

Suppose $p$ is the smallest prime dividing $n\in\mathbb N$ and suppose $kn+ap=m!$, where $ap<kn$, for some $a,k,m\in\mathbb N$, then there are $k^{\prime},a^{\prime}\in\mathbb N$ such that $k^{\prime}n+a^{\prime}p=(m+1)!$, where $a^{\prime}p<k^{\prime}n$.

Taking $k'=(m+1)k$ and $a'=a(m+1)$ works since $$k'n+a'p=(m+1)kn+a(m+1)p=(m+1)(kn+ap)=(m+1)!$$ and $$ap\lt kn\implies ap(m+1)\lt kn(m+1)\implies a'p\lt k'n$$