I would like to ask a probably simple question but I am not sure how to make it rigorous.
Let $p$ be a prime and $n$ a natural number whose lowest prime factor is $p$. If a natural number $x$ divides both $p!$ and $n$, is it true that $x=1$ or $p$?
I have no idea, but I am trying to say yes, and of course (if possible) elementarily without using named theorems except the fundamental theorem.
I am sure $x$ cannot be $2,3,...,p-1$ as it divides $n$, and also no prime factor of $x$ is lower than $p$. However, $x$ dividing $p!$ means that $x$ cannot have any prime factor greater than $p$ and also $x$ cannot be $p^k$ for every $k\geq 2$. Hence it is done by the fundamental theorem of arithmetics.
Is the above reasoning correct? Is something missing? And/or is something too much?
Thanks for the guide.