TITLE: Is this proof that $100!$ is not divisible by $101$ correct?
QUESTION [7 upvotes]: Is it enough to say that because $101$ is a prime number and $100!$ consists of numbers, and each of these numbers is composed of prime numbers which are less than $101$, and every number can't be decomposed in two different ways such that product of different primes is equal to another product of second group of a different primes, therefore $100!$ is not divisible by $101$?
Is this enough as a proof?

REPLY [5 votes]: You have the right idea. Here is a simpler argument based on your idea:
If $n \le 100$ and $p$ is a prime factor of $n$, then $p \le n \le 100$. Therefore, the primes in the factorization of $100!$ are precisely the primes less than $100$ and so $101$ is not one of them.