Proving that $n!$ divides $a(a+1)(a+2)...(a+n-1)$

Question

I need to prove that $n!$ always divides $S=a(a+1)(a+2)…(a+n-1)$.

I can see that developing the expression of $S$ will lead me towards $(a+1)^n$ which is the sum of $\sum_{0}^n\binom{n}{k}a^k$ which is obviously divisible by $n!$

My problem is that I don't see how can I prove it. If you could give me at least a hint, that would be great.

@J.W.Tanner I didn't think of that kind of approach, because the chapter is about combinatorics. But I don't really see where that one would lead... — Dknot, Apr 01 '20 at 10:02

score 6 · Accepted Answer · answered Apr 01 '20 at 09:55

6

$$a(a+1)(a+2)…(a+n-1)=\binom{a+n-1}{n}n!$$

Because $\binom{a+n-1}{n}\in\mathbb{Z}$ , $\; \; n!\;|\;a(a+1)(a+2)…(a+n-1)$

answered Apr 01 '20 at 09:55

Taha Direk

1,220

1

I didn't think of that kind of an approach. Thank you. – Dknot Apr 01 '20 at 10:08
@Dknot Not at all, all the best! – Taha Direk Apr 01 '20 at 10:17

score 1 · Answer 2 · answered Apr 01 '20 at 09:57

1

Lemma: The product of any $n$ consecutive positive integers is divisible by $n!$.

Proof:

Direct proof that $n!$ divides $(n+1)(n+2)\cdots(2n)$

answered Apr 01 '20 at 09:57

Dietrich Burde

130,978

Proving that $n!$ divides $a(a+1)(a+2)...(a+n-1)$

2 Answers2