Dividing Divisive Divisors Given a positive integer \$n\$ you can always find a tuple \$(k_1,k_2,...,k_m)\$ of integers \$k_i \geqslant 2\$ such that \$k_1 \cdot k_2 \cdot ... \cdot k_m = n\$ and $$k_1 | k_2 \text{ , } k_2 | k_3 \text{ , } \ldots \text{ , }k_{m-1}|k_m.$$
Here \$a|b\$ means \$b\$ is a multiple of \$a\$, say &quot;a divides b&quot;. If \$n&gt;1\$ all entries \$k_i\$ must be at least \$2\$. For \$n=1\$ we have no such factor and therefore we get an empty tuple.
In case you're curious where this comes from: This decomposition is known as invariant factor decomposition in number theory and it is used in the classification of finitely generated
Abelian groups.
Challenge
Given \$n\$ output all such tuples \$(k_1,k_2,...,k_m)\$ for the given \$n\$ exactly once, in whatever order you like. The standard sequence output formats are allowed.
Examples
  1: () (empty tuple)
  2: (2)
  3: (3)
  4: (2,2), (4)
  5: (5)
  6: (6)
  7: (7)
  8: (2,2,2), (2,4), (8)
  9: (3,3), (9)
 10: (10)
 11: (11)
 12: (2,6), (12)
108: (2,54), (3,3,12), (3,6,6), (3,36), (6,18), (108)

Related: http://oeis.org/A000688, List all multiplicative partitions of n
Haskell, 66 62 60 bytes

f n=[n|n&gt;1]:[k:l:m|k&lt;-[2..n],l:m&lt;-f$div n k,mod(gcd n l)k&lt;1]


Try it online!