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I am trying to proof that for $n,m\in\mathbb{N}$ with $m\leq n$ the binomial coefficient $\binom{n}{m}=\frac{n!}{m!(n-m)!}$ is always a natural number. For I need to prove that $m!(n-m)!|n!$ for every $n,m\in\mathbb{N}$ with $m\leq n$.

Now, obviously the statement is true for $n=0$. Yet, once I assume that the statement is true for some $n$ I can't make proof that it holds for $n+1$. Here is where I got stuck: \begin{eqnarray} (n+1)! &=& (n+1)n!=(n+1-m)n! + n!m\\ &=& (n+1-m)!\frac{n!}{(n-m)!}+n!m\\ &=& (n+1-m)!\frac{km!(n-m)!}{(n-m)!}+n!m\\ &=& km!(n+1-m)!+n!m\\ \end{eqnarray} If anyone completed the proof by induction, I would be very thankful.

EDIT - I now have proven the statement this way using $\binom{n+1}{m}=\binom{n}{m-1}+\binom{n}{m}$:

$$\binom{n+1}{m}=\binom{n}{m-1}+\binom{n}{m}=k_0+k_1$$ with $k_0, k_1\in\mathbb{N}$, following our assumption. It follows that $$(n+1)!=(k_0+k_1)m!(n+1-m)!$$ thus proving the statement for $1\leq m\leq n$. The cases left are easy and I am omiting them here.

Still, I would like to see a prove by induction that utilizes the statement $(n+1)!=(n+1-m)n!+mn!$.

EDIT 2 - This is not a duplicate of this question, since, although the statement to proof is the same, what I am asking for is a proof by induction that uses a specific property of $(n+1)!$.

The proof I gave in my first edit was based on the second most up-voted answer given in the other post, but it is not what I am asking for.

As for the "best" answer given in the article this is supposedly a duplicate of: Neither that nor the answer it links to in another article constitute neither a proof of the statement, and much less a proof by induction.

Watson
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    Well, it is not a question either! – barak manos Sep 04 '16 at 14:56
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    And this is not a comment! – Patrick Stevens Sep 04 '16 at 14:57
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    I think what you want just might not be really feasible. There is a reason why the proofs you found do not proceed by straightforward induction. But alright, in this form it may not be a duplicate. – quid Sep 04 '16 at 15:09
  • Why would anyone want a proof by induction over a direct proof? Never mind. If indeed you don't want a general proof but a specific proof you should modify your title and first paragraph to reflect that. – fleablood Sep 04 '16 at 15:36
  • @TahaAkbari the question was closed as dupe of this. The edits say it is not a dupe and so it was reopened. Now, you might be aware of this yet still of a different opinion, but if this is the case then please explain it. – quid Sep 04 '16 at 15:46

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