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Prove that the product of the $n$ integers of an arithmetic progression of $n$ terms is divisible by $n!$ if common difference is relatively prime to $n!$. First part of the question was to prove for d=1 which I was able to do using binomial coefficient. In this question though I am absolutely stumped.

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Assume that the result is true for $d=1$, so $n!$ divides $(a+1)\cdots(a+n)$ for any $a$. In modular mathematics, this is written as $(a+1)\cdots(a+n)\equiv 0\pmod{n!}$.

If $d$ is coprime to $n!$, then it is invertible, and has some inverse $c$, $cd\equiv 1\pmod{n!}$. So $$(a+d)(a+2d)\cdots(a+nd)\equiv d^n(ac+1)(ac+2)\cdots(ac+n)\equiv 0\pmod{n!}$$

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