An eely function $\mu (n):\;\;\prod\limits_{k = 0}^{n - 1} {\left( {\mu (n) - \mu (k)} \right)} = 1$

This is only a partial answer.

If your claim is correct, i.e. differences are decreasing, then the sequence is sublogarithmic. Indeed, let $s(n) := \mu(n+1) -\mu(n) $. The recurrence relation, using the claim $s(j) > s(n)$ for $j<n$, rewrites as

$$ 1 = \prod_{k=0}^n (\sum_{j=k}^n s(j) ) > \prod_{k=0}^n (n-k+1) s(n) = (n+1)! s(n) ^{n+1} $$

Which implies that

$$ s(n) < \left ( \frac{1}{(n+1)! } \right ) ^{1/(n+1) }$$

The Stirling approximation can be written as $m! \approx (m/e) ^{m}$. Substituting we get that $s(n) < \frac{e}{n+1}$ . Thus we have

$$ \mu(n) = s(n-1) +\ldots+s(0) < $$ $$ e \sum_{k=0}^{n-1} \frac{1}{k+1} \approx e \log(n) $$

We also have a sort of converse to this. Suppose that for all n we have $$(*) \ \ s(n) > \left ( \frac{1}{(n+2)! } \right ) ^{1/(n+2) }$$

That is, the sequence is also superlogarithmic. We want to show that differences are decreasing. Suppose that at some point $n$ we have for the first time that the difference increase, aka $s(n) \ge s(n-1) $. Then we get

$$ 1 = \prod_{k=0}^n (\sum_{j=k}^n s(j) ) \ge \prod_{k=0}^n (n-k+1) s(n-1) = (n+1)! s(n-1) ^{n+1} $$

From which we get, using again Stirling approximation, the opposite (*) inequality, yielding a contradiction.

This argument can be made precise with inequalities instead of plain Stirling approximation, but since it is not a strong result with respect to your requests, I think it's not worth. Hope it can inspire someone else for a more sophisticated approach :)

Edit. I have an approach for the lower bound. By substituting each factor with the biggest $s$ appearing we get a reversed inequality

$$ 1\le n s(0) \cdot (n-1)s(1) \ldots 1 \cdot s(n-1) $$

Rearranging terms and taking the logarithms we get

$$ 0 \le \log ns(n-1) + \ldots +\log 1\cdot s(0) $$

Let us define $$\sigma(n) := \frac{1}{n} \sum_{j=0} ^{n-1} \log (j+1) s(j) $$ Equation above ensures that $\limsup \sigma(n) \ge 0$. Now the plot twist: by Stolz Cesaro the limsup of the general term is greater or equal than the average one, so that

$$ \limsup \log s(n) (n+1) \ge \limsup \sigma(n) \ge 0$$

Which implies that

$$ \limsup s(n) (n+1) \ge 1$$

Which provides a lower bound like estimate. If this can be refined to be a limit, than by summing over n we would get that $\mu(n) \ge \log(n) $ asymptilotically.