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In response to a request by @Ari for a proof by induction that the product of $n$ consecutive integers is divisible by $n!$, I attempted what I thought was an argument more plain and direct. Taking a product of $n$ consecutive integers $k(k+1)(k+2)...(k+n-1)$, then since $1$ divides $k$, and $2$ divides $k$ or $(k+1)$, and $3$ divides $k$ or $k+1$ or $k+2$...and $n$ divides either $k$ or $k+1$ or $k+2$...or $k+n-1$, therefore $n!$ divides $k(k+1)(k+2)...(k+n-1)$. But I was brought by @JMoravitz to see that the conclusion does not follow. For taking $6!$ and $11\cdot13\cdot17\cdot4\cdot5\cdot6$, every integer $\le6$ divides $11, 13, 17, 4, 5,$ or $6$, but $6!$ does not divide $291720$. This is of course because $6!$ contains $2^4$ and $3^2$, while $291720$ contains only $2^3$ and $3^1$. The proof thus requires the fulfillment of a second condition, namely that the product of $n$ integers contain all of its prime factors at least as many times as $n!$ does. This seems to be assured by the fact that the $n$ integers are consecutive. If so, what is a good way to prove it?

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    $\begingroup$ Link to the original question that OP is referring to. The double-induction proof which I alluded to in the other post still seems strongest and I do not see a convenient way around it yet, especially with both $n$ and $k$ varying. Regardless, I'll give someone else the chance to try explaining this time. $\endgroup$ – JMoravitz Jan 8 '17 at 0:14
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    $\begingroup$ Well, there's always the proof by binomial coefficients. $\binom {k+n-1}n=\frac {(k+n-1)!}{n! (k-1)!}=\frac {k(k+1)\cdots (k+n-1)}{n!}$ and of course the binomial coefficient is always an integer. $\endgroup$ – lulu Jan 8 '17 at 0:22
  • $\begingroup$ @lulu One could argue that, at least without the presence of combinatorial arguments, this is circular logic. The given question here is exactly what you need to prove if you want to prove arithmetically / algebraically that the binomial coefficients are integers. $\endgroup$ – Arthur Jan 8 '17 at 1:59
  • $\begingroup$ @Arthur But of course you do have the combinatorial arguments. The quotient is the number of ways of choosing $n$ objects out of $k+n-1$. I entirely agree that this is critical to the argument. $\endgroup$ – lulu Jan 8 '17 at 11:19
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As shown in equation $(3)$ of this answer, the number of factors of $p$ in $n!$ is given by Legendre's Formula: $$ v_p(n!)=\frac{n-\sigma_p(n)}{p-1}\tag{1} $$ where $\sigma_p(n)$ is the sum of the digits of the base-$p$ representation of $n$. Therefore, the number of factors of $p$ in $\frac{(n+k)!}{k!}$ is $$ v_p\left(\frac{(n+k)!}{k!}\right)=\frac{n-\sigma_p(n+k)+\sigma_p(k)}{p-1}\tag{2} $$ The difference of $(1)$ and $(2)$ is $$ v_p\left(\frac{(n+k)!}{k!}\right)-v_p(n!)=\frac{\sigma_p(n)+\sigma_p(k)-\sigma_p(n+k)}{p-1}\tag{3} $$ and $(3)$ is the number of carries when adding $n$ and $k$, and therefore, greater than or equal to $0$.

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