Number of factorizations of distinct factors Let $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ an integer with $p_i$ prime and $e_i \in \mathbb N$. The prime factorization can assumed to be known, i.e., we already know $p_1, \dotsc, p_k$ and $e_1, \dotsc e_k$.
Is it possible to find the number of factorizations of length $m$ of the form 
$n = n_1 \cdot n_2 \dotsm n_m$ such that $n_1 < n_2 < \dotsb < n_m$ other than brute forcing?
(That means two factorizations that are just a rearrangement of each other are counted as the same one, e.g. $1 \times 2 \times 3$ is considered the same  as $1 \times 3 \times 2$, so we only count those in ascending order.)
Example: For the number $n = 12 = 2^2 \cdot 3$ we have following factorizations with the factors in ascending order:
For $m=1$ we have one:


*

*$12$


For $m=2$ we have three:


*

*$1 \times 12$

*$2 \times 6$

*$3 \times 4$


For $m=3$ we have two:


*

*$1 \times 2 \times 6$

*$1 \times 3 \times 4$

 A: Let  me present  a proof  of the  first special  case (product  of $k$
distinct primes) by @martin, which is a nice result that can be proved
by Polya enumeration.

I will assume the reader  has consulted and understood the material at
the following
MSE link 
which I will not duplicate here.

Using the notation from the link  with $q$ being the number of factors
in  the partition  we  obtain  by the  Polya  Enumeration Theorem  the
following formula:
$$G(k, q) = \left[\prod_p X_p\right]
Z(P_q)\left(\prod_p (1+X_p)\right)
\quad\text{where}\quad n=\prod_p p^v$$
with all  $v=1$ and we have  $k$ distinct primes in  the product. Here
the  square bracket  denotes  coefficient extraction  of formal  power
series  and  $Z(P_q)$   is  the  cycle  index  of   the  set  operator
$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}_{=q}$ which  was also used in  the linked-to computation from above.
Now recall the OGF of the set operator which is
$$Z(P_q) = [z^q]
\exp\left(a_1 z - a_2 \frac{z^2}{2}
+ a_3 \frac{z^3}{3}
- a_4 \frac{z^4}{4}
+\cdots \right).$$
Observe that on substituting into the cycle index we let
$$a_m = \prod_p (1+X_p^m).$$
But the  degree of  $X_p$ in the  coefficient being extracted  is one,
which means that  from the $a_m$ with $m\ge 2$  only the constant term
contributes, which is one.

This gives the formula
$$G(k, q) = \left[\prod_p X_p\right] [z^q]
\exp\left(z\prod_p (1+X_p) 
- \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots\right)$$
which is
$$G(k, q) = \left[\prod_p X_p\right] [z^q]
\exp\left(z\left(-1+\prod_p (1+X_p)\right) + \log(1+z)\right)
\\ = \left[\prod_p X_p\right] [z^q] (1+z) 
\exp\left(z\left(-1+\prod_p (1+X_p)\right)\right)
\\ = \left[\prod_p X_p\right]
\left(\frac{1}{q!} \left(-1+\prod_p (1+X_p)\right)^q
+ \frac{1}{(q-1)!} \left(-1+\prod_p (1+X_p)\right)^{q-1}\right).$$
Doing coefficient extraction on the first term
we find
$$\left[\prod_p X_p\right]
\frac{1}{q!} \sum_{m=0}^q {q\choose m} (-1)^{q-m}
\prod_p (1+X_p)^m.$$
Only the terms with $X_p$ raised to the power one contribute
and we get for the first term
$$\frac{1}{q!} \sum_{m=0}^q {q\choose m} (-1)^{q-m} m^k
= {k\brace q}.$$
The second  term is  similar and therefore  the answer to  the special
case of a product of $k$ primes is
$${k\brace q} + {k\brace q-1}.$$
Now  that   we  have   this  we  can   easily  give   a  combinatorial
interpretation. The first term represents the case where we divide the
$k$ prime factors into $q$  non-empty sets which correspond to the $q$
distinct  factors of  the multiplicative  partition with  none  of the
factors being one. (With the $k$ primes being distinct the products of
the  elements of these  sets are  necessarily distinct.)   This almost
completes  the  count except  we  have  not  accounted for  partitions
containing  one as  a factor.  That leaves  $q-1$ distinct  factors to
choose according to the same procedure as before, done.

Remark. Using the OGF of the Stirling numbers of the second kind
which is
$${n\brace k} = [z^n] \prod_{r=1}^k \frac{z}{1-rz}$$
we get the generating function
$$\prod_{r=1}^q \frac{z}{1-rz}+
\prod_{r=1}^{q-1} \frac{z}{1-rz}
= \left(1+\frac{z}{1-qz}\right) \prod_{r=1}^{q-1} \frac{z}{1-rz}
\\ = \frac{1-(q-1)z}{1-qz}  \prod_{r=1}^{q-1} \frac{z}{1-rz}.$$
