Is is true that for all $n\in\mathbb{N}$ that $\prod\limits_{i=1}^{n}(a+i) \equiv 0 \ [n!]$?

Question

Let $a\in \mathbb{Z}$, we want to prove that for all $n\in\mathbb{N}$ that $\prod\limits_{i=1}^{n}(a+i) \equiv 0 \ [n!]$ ?

By induction :

For $n=1$ we obtain that $1!$ divides $a+1$.

For $n=2$ we obtain that $2!$ divides $(a+1)(a+2)$ because one term is necessary even.

For the case $n+1$ I need a hint.

Thanks in advance !

Answer 1

Note that at least for $a\in\Bbb N$, we have that ${a+n\choose a}$ is an integer. This can be extended to $a\in \Bbb Z$.