A problem involving the product $\prod_{k=1}^{n} k^{\mu(k)}$, where $\mu$ denotes the Möbius function Let $\mu$ denote the Möbius function whereby
$$\mu(k)    = 
  \begin{cases} 
   0 & \text{if $k$ has one or more repeated prime factors} \\
   1 & \text{if $k=1$} \\
   (-1)^j       & \text{if $k$ is a product of $j$ distinct primes}\end{cases}$$ for all $k \in \mathbb{N}$. I have previously noted a surprising connection between products of the form $$\prod_{k=1}^{n} k^{\mu(k)}$$ and harmonic numbers (see http://oeis.org/A130087). Letting $H_{n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$ denote the $n^{\text{th}}$ harmonic number, I noted that the denominator (in lowest terms) of this product is equal to the denominator (in lowest terms) of $H_{n}^{2}n!$ for all $n < 897$.  For example, letting $n = 11$, we have that $$1^{\mu(1)} \cdot 2^{\mu(2)} \cdot \cdots \cdot 11^{\mu(11)} = \frac{2}{77},$$ and $$H_{11}^{2}11! = \frac{28030126084}{77}.$$ This property does not hold for all $n \in \mathbb{N}$, and the first counterexample occurs in the case whereby $n = 897$. Letting $\text{den}(q)$ denote the denominator in lowest terms of $q \in \mathbb{Q}_{> 0}$, it is not obvious to me why the equality $$\text{den}\left(\prod_{k=1}^{n} k^{\mu(k)}\right) = \text{den}\left(H_{n}^{2}n!\right)$$ holds for the first several hundred natural numbers. So it seems natural to me to ask:
(1) Is there some simple number-theoretic explanation as to 'why' this equality holds for the first several hundred natural numbers?
(2) Is there some simple number-theoretic interpretation as to 'why' this equality does not hold in general (which specifically explains the counterexample in the case whereby $n = 897)$?
(3) Is there a natural combinatorial way of approaching this problem? Note that integers of the form $H_{n}n!$ are unsigned Stirling numbers of first kind, so it is plausible that there is a natural combinatorial interpretation of the expression $\text{den}\left(H_{n}^{2}n!\right)$.
 A: Let $k_p(n)$ be defined as the non-negative integer such that $p^{k_p(n)}≤n$ but $p^{k_p(n)+1}>n$. Furthermore let $\epsilon_p(n)$ be the exponent of $p$ in the prime factorisation of $n!$; it is well known that $\epsilon_p(n)=\sum_{k=1}^{\infty}\lfloor\frac{n}{p^k}\rfloor$. Define $\mathcal{P}_n:=\{p \text{ prime } : \frac{n}{2}<p≤n\}$.
We are going to show:
$$ \bbox[border:2px solid red]
{
\text{den}\left(\left(H_n\right)^2\cdot n!\right)=\prod_{p\in\mathcal{P}_n}p
}
$$
For $n≥6$.
Firstly, we note that
$$
(H_n)^2n!=\\
\frac{n!}{(\text{lcm}1,...,n)^2}\left(\sum_{k=1}^{n}\frac{\text{lcm}(1,...,n)}{k}\right)^2
$$
So lets calculate:
$$
\frac{\text{lcm}(1,...,n)^2}{\text{gcd}\left(\text{lcm}(1,...,n)^2, n!\right)}
$$
With the notation introduced above, we obtain $\text{lcm}(1,...,n)=\prod_{p\text{ prime}}p^{k_p(n)}$ and $n!=\prod_{p\text{ prime}}p^{\epsilon_p(n)}$ and thus:
$$
\text{gcd}\left(\text{lcm}(1,...,n)^2, n!\right)=\prod_{p\text{ prime}}p^{\min\left(2 k_p(n)\ ,\ \epsilon_p(n)\right)}
$$
Furthermore $\epsilon_p(n)=\sum_{k=1}^{k_p(n)}\lfloor\frac{n}{p^k}\rfloor$. We distinguish the following cases:
case 1: $\lfloor\frac{n}{p^{k_p(n)}}\rfloor≥2$
$$
\epsilon_p(n)=\sum_{k=1}^{k_p(n)}\lfloor\frac{n}{p^k}\rfloor≥\sum_{k=1}^{k_p(n)}2=2k_p(n)
$$
case 2: $\lfloor\frac{n}{p^{k_p(n)}}\rfloor=1$ and $k_p(n)≥3$
$$
\epsilon_p(n)=\sum_{k=1}^{k_p(n)}\lfloor\frac{n}{p^k}\rfloor≥\sum_{k=1}^{k_p(n)}\lfloor\frac{p^{k_p(n)}}{p^k}\rfloor=\sum_{k=0}^{k_p(n)-1}p^k\overset{AM-GM}{≥}k_p(n)\left[\prod_{k=0}^{k_p(n)-1}p^k\right]^{\frac{1}{k_p(n)}}=k_p(n)\cdot p^{\frac{k_p(n)-1}{2}}≥2k_p(n)
$$
case 3: $\lfloor\frac{n}{p^{k_p(n)}}\rfloor=1$ and $k_p(n)=2$
$$
\epsilon_p(n)=\lfloor\frac{n}{p}\rfloor+1
$$
Since $\lfloor\frac{n}{p^2}\rfloor=1$ we have $p^2≤n<2p^2$. Thus, if $p≥3$ we have $\frac{n}{p}≥p≥3\implies \lfloor\frac{n}{p}\rfloor≥3\implies \epsilon_p(n)≥4=2k_p(n)$ and if $p=2$, we simply calculate the cases where $n<8$; we see that the formula holds whenever $n\neq4$ and $n\neq5$.
case 4: $\lfloor\frac{n}{p^{k_p(n)}}\rfloor=1$ and $k_p(n)=1$
$$
\epsilon_p(n)=\lfloor\frac{n}{p}\rfloor=1<2=2k_p(n)
$$
Here we have the only case where $\epsilon_p(n)<2k_p(n)$; so this inequality is true iff $p≤n<2p \iff p\in\mathcal{P}_n$
If we combine the four cases, we see that:
$$
\text{gcd}\left(\text{lcm}(1,...,n)^2, n!\right)=\prod_{p\text{ prime}}p^{\min\left(2 k_p(n)\ ,\ \epsilon_p(n)\right)}=\left(\prod_{p\notin\mathcal{P}_n\ ,\ p\text{ prime}}p^{2k_p(n)}\right)\cdot\left(\prod_{p\in\mathcal{P}_n}p\right)
$$
And thus:
$$
\text{den}\left(\frac{n!}{\text{lcm}(1,...,n)^2}\right)=\frac{\text{lcm}(1,...,n)^2}{\text{gcd}\left(\text{lcm}(1,...,n)^2, n!\right)}=\frac{\prod_{p\text{ prime}}p^{2k_p(n)}}{\left(\prod_{p\notin\mathcal{P}_n\ ,\ p\text{ prime}}p^{2k_p(n)}\right)\cdot\left(\prod_{p\in\mathcal{P}_n}p\right)}=\frac{\left(\prod_{p\in\mathcal{P}_n}p^2\right)}{\left(\prod_{p\in\mathcal{P}_n}p\right)}=\prod_{p\in\mathcal{P}_n}p
$$
Furthermore, we see that for $p\in\mathcal{P}_n$ the only summand in $\sum_{k=1}^n\frac{\text{lcm}(1,...,n)}{k}$ which isn't divisible by $p$ is the one with $p$ in the denominator i.e. $\frac{\text{lcm}(1,...,n)}{p}$ so the sum isn't divisible by $p$. Therefore we obtain:
$$
\text{den}\left((H_n)^2n!\right)=\text{den}\left(\frac{n!}{\text{lcm}(1,...,n)^2}\right)=\prod_{p\in\mathcal{P}_n}p
$$
Whenever $n\neq4,5$.
This isn't an inherently combinatorial identity, as you wished, but it breaks it down quite a bit. However, I fail to see how to justify your observations with this identity, but it sheds new light on what we really have to prove:
If $p\in\mathcal{P}_n$ then the prime $p$ appears only once in the factors of the numbers $\{1,...,n\}$. As $\mu(p)=-1$ it appears in the denominator of $\prod_{k=1}^{n}k^{\mu(k)}$. As it can't be cancelled out, we can see that 
$$\prod_{p\in\mathcal{P}_n}p\ |\ \text{den}\left(\prod_{k=1}^{n}k^{\mu(k)}\right)$$
In order to have equality, we would need to have that every prime not in $\mathcal{P}_n$ appears more ore equally often in the numerator than in the denominator. For $n=897$ this fails for $p=23$; it appears one time more often in the denominator than in the numerator and we then have $$\prod_{p\in\mathcal{P}_{897}}p=23\cdot\text{den}\left(\prod_{k=1}^{897}k^{\mu(k)}\right)$$
I doubt that I will come up with an explanation and characterization for the cases of equality, but I will continue to work on it.
Edit: Some further results:
We're going to find an expression for $h_p(n)$, the exponent of $p$ in the denominator of $\prod_{k=1}^n k^{\mu(k)}$ in lowest terms (negative if it's in the numerator and $0$ if it is cancelled out). We have:
$$
h_p(n)=\sum_{k\leq n\ ,\ p|k}-\mu(k)=\sum_{k\leq \frac{n}{p}}-\mu(pk)=\sum_{k\leq \frac{n}{p}}\mu(k)+\sum_{k\leq \frac{n}{p}\ ,\ p|k}-\mu(k)=M\left(\frac{n}{p}\right)+h_p\left(\lfloor\frac{n}{p}\rfloor\right)
$$
Where $M$ is the Mertens function defined to be $M(x):=\sum_{1\leq k\leq x}\mu(k)$. So inductively we obtain:
$$
h_p(n)=\sum_{k=1}^{\infty}M\left(\frac{n}{p^k}\right)
$$
In a nice analogy to $\epsilon_p(n)$. This gives us a family of counter examples to your observation:
Let $m\geq 6$ (our formula for the harmonic denominator needs to be true) be such that $M(m)\geq 0$ (I think there are infinitely many such $m$) and let $p\in\mathcal{P}_m$, which exists due to Bertrand's postulate. Let $n=pm+r$ such that $0\leq r <p$. So we verify that $\lfloor\frac{n}{p}\rfloor=m$ and $\lfloor\frac{n}{p^2}\rfloor=1$ (because $r<p\leq p(2p-m)\implies \frac{n}{p^2}=\frac{mp+r}{p^2}<2$) and therefore:
$$
h_p(n)=M(m)+M(1)\geq M(1)=1
$$
But $2p\leq p^2\leq pm\leq n$ an thus $p\notin \mathcal{P}_n$ and we obtain that 
$$
\prod_{p\in\mathcal{P}_{n}}p<\text{den}\left(\prod_{k=1}^{n}k^{\mu(k)}\right)
$$
So $n$ is a counter example. This also suggests $n=897$ to be the first one: the smallest $m>5$ with $M(m)\geq 0$ is $m=39$ and the smallest prime of $\mathcal{P}_{39}$ is $23$. With $r=0$ we obtain $n=897$ and from there on we find many counter examples.
