Prove $n!\mid\prod_{k=i}^{i+n-1}k$

I have no idea how to prove this, I haven't yet learned much about this kind of product.

$$n!\mid\prod_{k=i}^{i+n-1}k$$

• First try proving the product of j consecutive integers is divisible by j. – user222031 Mar 30 '15 at 21:35

Hint a very interesting fact you can use :

$$\dbinom {i+n-1} {i-1}=\frac{\cdots }{n!}$$

• Could you expand on that? I got to $\frac{i+n-1}{n!}$, but what can I use that for? – YoTengoUnLCD Mar 30 '15 at 22:00
• if you take $p=\prod_{k=i}^{i+n-1}k$ then $$\frac{p}{n!}=\dbinom {i+n-1} {i-1}$$ and we know that binomial coefficients are integers, hence $\frac{p}{n!}$ is an integer which signifies that $n!$ divides $p$ – Elaqqad Mar 30 '15 at 22:03
• What if $i+n-1 < 0$? then I shouldn't be able to use the binomial coefficient identity, right? – YoTengoUnLCD Mar 30 '15 at 22:09
• directly you can't, but you can do it separably , by replacing every term $k$ by $-k$ and the product will become a product of consecutive positive integers – Elaqqad Mar 30 '15 at 22:14

Notice that this relates strongly to the so-called choose function, $C(n,k)$, which is defined as $$C(N,k) = \frac{N!}{k!(N-k)!}$$ Since the right-hand side can be written $\prod_{k=i}^{i+n-1} k = \frac{(i+n-1)!}{(i-1)!}$, the question is really asking you to verify that $$C(n+i-1, n) = \frac{(i+n-1)!}{n!(i-1)!} \in \mathbb Z$$

• I realized I could write the product that way, but I don't understand what to do next, how can I approach verifying that? – YoTengoUnLCD Mar 30 '15 at 21:46
• Also, doesn't this only work if $i>1$? – YoTengoUnLCD Mar 30 '15 at 22:01