Is this proof that $100!$ is not divisible by $101$ correct?

Question

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?

Answer 1

By Wilson's theorem , $100!=(101-1)!\equiv -1\pmod{101}$, since $101$ is prime.

Answer 2

I think the argument relies hazily on the uniqueness of prime factorisation. It would be much more irrefutable to use

Euclid's lemma : If a prime number divides a product of numbers, it divides at least one factor in the product.

Answer 3

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.