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

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

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  • $\begingroup$ First try proving the product of j consecutive integers is divisible by j. $\endgroup$ – user222031 Mar 30 '15 at 21:35
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Hint a very interesting fact you can use :

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

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  • $\begingroup$ Could you expand on that? I got to $\frac{i+n-1}{n!}$, but what can I use that for? $\endgroup$ – YoTengoUnLCD Mar 30 '15 at 22:00
  • $\begingroup$ 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 $p$ $\endgroup$ – Elaqqad Mar 30 '15 at 22:03
  • $\begingroup$ What if $i+n-1 < 0$? then I shouldn't be able to use the binomial coefficient identity, right? $\endgroup$ – YoTengoUnLCD Mar 30 '15 at 22:09
  • $\begingroup$ 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 $\endgroup$ – Elaqqad Mar 30 '15 at 22:14
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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 $$C(n+i-1, n) = \frac{(i+n-1)!}{n!(i-1)!} \in \mathbb Z$$

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

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