Prove that $6! \mid n(n+1)....(n+5)$

Question

Prove that for all $n \in \mathbb{Z}$, $6! \mid n\cdot(n+1)\cdots(n+5)$ using only criteria of divisibility (without using combinatorial arguments).

Answer 1

Hint: $n$, $(n+1)$, $(n+2)$, $(n+3)$, $(n+4)$, $(n+5)$ are all 6 consecutive numbers so one of them is a multiple of 6, and 3 of them are a multiple of 2; and 2 of them are a multiple of 3...

Answer 2

Consider the primes that $6!=2^4\cdot 3^2\cdot 5$ has.

Since $A=n(n+1)(n+2)(n+3)(n+4)(n+5)$ is a product of six consecutive numbers, there are three even numbers, and there is at least one number divisible by $4$.

There are two numbers divisible by $3$. Also, there exists a number divisible by $5$.

Hence, $A$ is divisible by $2^4\cdot 3^2\cdot 5=6!.$