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.
