Prove that $(a+1)(a+2)(a+3)\cdots(a+n)$ is divisible by $n!$

Question

so I have this math problem, I have to prove that $$(a+1)(a+2)(a+3)\cdots(a+n)\text{ is divisible by }n!$$ I'm not sure how to start this problem... I completely lost. Here's what I know: $(a+1)(a+2)(a+3)\cdots(a+n)$ is like a factorial in that we are multiplying consecutive terms, $n$ times. However, how would I prove that it is divisible by $n$? Thanks

Answer 1

Let $$ f(a,n) = (a+1)(a+2)\cdot\ldots\cdot(a+n).\tag{1}$$ We have: $$ f(a+1,n)-f(a,n) = n\cdot f(a+1,n-1) \tag{2} $$ hence the claim follows by applying a double induction.

Answer 2

$$(a+1)(a+2)\dots(a+n) = \frac{(a+n)!}{a!}$$

We know $$\frac{(a+n)!}{(a!)(n!)} = {a+n\choose n} \in \mathbb{N}$$

So, $n!$ divides $\frac{(a+n)!}{a!} = (a+1)(a+2)\dots(a+n)$