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This question already has an answer here:

Can $n!$ be the product of $k$ consecutive integers for $k > 1$? (Not including the degenerate cases such as when $k = 2$, then $1\cdot2 = 2!$ and $2\cdot 3 = 3!$, and so on.)

I am asking not for $n!$ to be divisible by $n$ consecutive integers, I am asking for $n!$ to equal the product of $k$ consecutive integers, implying that $n$ is not necessarily equal to $k$ (And when $n = k$, then clearly there exists an answer, namely $n! = (1)(2)(3)...(k)$)

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marked as duplicate by Steven Stadnicki, Daniel Fischer, Dan Rust, Davide Giraudo, Brandon Carter Sep 2 '13 at 19:39

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    $\begingroup$ $5! = 4\cdot 5\cdot 6$? $6! = 8\cdot 9\cdot 10$. $\endgroup$ – Daniel Fischer Sep 2 '13 at 18:22
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    $\begingroup$ $n!=1\cdot 2\cdot 3\cdot ...\cdot n$ is defined so as to be the product of $n$ consecutive integers. $\endgroup$ – walcher Sep 2 '13 at 18:22
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    $\begingroup$ $n!$ can be the product of $n$ consecutive integers as well as $n-1$ consecutive integers. $\endgroup$ – Tomas Sep 2 '13 at 18:22
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    $\begingroup$ Also $(k-1)!$ can be written as product of $k!-k$ integers in the following way: $(k+1)\cdot\dots\cdot k!$. $\endgroup$ – Tomas Sep 2 '13 at 18:26
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    $\begingroup$ 6! = 720 can be written as the product of three other consecutive integers: 8*9*10 Is that what you mean? $\endgroup$ – imranfat Sep 2 '13 at 18:38
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The formula $$(n!-1)!=\frac{(n!)!}{n!}=\prod_{k=n+1}^{n!} k$$ gives an infinite family of examples.

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