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As in the title, we have the product $$\prod_{k=1}^n \left(\prod_{j=1}^k\frac{k}{j}\right)$$ for which we want to show that it is integer for every $n \in \mathbb{N}$ (with $n > 0$).

So far I have gotten rid of the inner product sign, and gotten $$\prod_{k=1}^n \frac{k^k}{k!}$$ Now, each $a$ occurs $n + 1 - a$ times in the denominator, so we can also write that as $$\prod_{k=1}^n \frac{k^k}{k^{n+1-k}} = \prod_{k=1}^n \frac{k^{2k}}{k^{n+1}}=\frac{\prod_{k=1}^n k^{2k}}{\left(n!\right)^{n+1}}$$ Now I tried arguing that every (maybe every prime) $p$ occurs in the denominator $a(n+1)$ times and $2p + 4p + ...+ 2ap = pa(a+1)$ times in the denominator, when $a=\lfloor\frac{n}{p}\rfloor$, and that, when comparing these two, we find that $pa(a+1) \geq a(n+1)$, since $p(a+1) > n$, and that therefore each factor should occur more often in the numerator than the denominator. However, I think this faces some problems when some powers of $p$ are also smaller than $n$, as these also have to be canceled out, but are already used to cancel $p$ itsself.

Hence, my question is how to complete this way of proving the assertion and how to deal with the power problem, or how to find a different solution.

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Note that $$\prod_{k=1}^{n} \binom{n}{k}=\prod_{k=1}^{n} \frac{n^{\underline{k}}}{k!}=\prod_{k=1}^n \frac{k^k}{k!}$$ As $$\prod_{k=1}^{n}n^{\underline{k}}=\prod_{k=1}^n k^k$$ This follows elementarily from following the terms.

NOTE:In this context, $n^{\underline{k}}$ signifies the falling factorial function. Be sure not to confuse it with $n^k$.

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Your start was good:

$$\prod_{k=1}^n \left(\prod_{j=1}^k\frac{k}{j}\right)$$

First simply convert the inner formula into a closed one, as you did.

$$\prod_{k=1}^n \frac{k^k}{k!}$$

From that point, the things are more easy as they seem. Here is the formula more clearly, without product:

$$\frac{1^1 \cdot 2^2 \cdot 3^3 \cdot ... \cdot n^n}{1! \cdot 2! \cdot ... \cdot n!}$$

We convert also the nominator to a factorial expression:

$1^1 \cdot 2^2 \cdot 3^3 \cdot ... \cdot n^n = \frac{n!}{1!} \cdot \frac{n!}{2!} \cdot ... \cdot \frac{n!}{n!}$

The result is the formula:

$\frac{\frac{n!}{1!} \cdot \frac{n!}{2!} \cdot ... \cdot \frac{n!}{n!}}{1! \cdot 2! \cdot ... \cdot n!}$

From that point, you can easily see that here are the product of the binomial constants:

$\frac{\frac{n!}{1!} \cdot \frac{n!}{2!} \cdot ... \cdot \frac{n!}{n!}}{n! \cdot (n-1)! \cdot ... \cdot 1!}$

$\frac{n!}{1! \cdot n!} \cdot \frac{n!}{2! \cdot (n-1)!} \cdot ... \cdot \frac{n!}{n! \cdot 1!}$

Q.E.D.

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