Is the exponential version of Catalan-Dickson conjecture true? The aliquot sum function $s:\mathbb{N}\rightarrow \mathbb{N}$ assigns to any natural number $n$ the sum of its proper divisors. Perfect numbers are fixed points of this function. The open conjecture of Catalan and Dickson predicts the behavior of $\{s^{(k)}(n)\}_{k\in \omega}$, the aliquot sequence of $n$ produced by the infinite iteration of $s$ on $n$. The conjecture states that for any natural number $n$, $\{s^{(k)}(n)\}_{k\in \omega}$ ends up becoming constant on $0$ or a perfect number or circulating in a loop of sociable numbers.
Now, let's assume an analogy of this conjecture for the exponentiation operator rather than the multiplication.
Definition. For natural numbers $n, c, d$:
(1) $c$ and $d$ are exp-divisors of $n$ if $c^d=n$. Any of them is called proper if it is not equal to $n$.
(2) The aliquot product function $t:\mathbb{N}^{>1}\rightarrow \mathbb{N}$ assigns to any natural number $n>1$ the product of its proper exp-divisors.
(3) The exp-aliquot sequence of $n$ is $\{t^{(k)}(n)\}_{k\in \omega}$.
(4) An exp-perfect number is a fixed point of $t$.
(5) A finite sequence of natural numbers is called exp-sociable if they form an aliquot product loop. (i.e. Any of them is the aliquot product of the previous one and the first number is the aliquot product of the last one).
Example. We have $3^2=9^1$ so $1, 2, 3$ are proper exp-divisors of $9$ and so $t(9)=2.3=6$. As $6$ has no proper exp-divisors except $1$ so $t^2(9)=t(6)=1$.
Remark. Note that just like the case of abundant numbers, there are natural numbers whose aliquot product is greater than the number itself. For instance, we have $2^8=4^4=16^2=256^1$. Thus the proper exp-divisors of $256$ are $1, 2, 4, 8, 16$ and so $t(256)=1024$. But even in this case the aliquot sequence eventually terminates at $1$ because we have $t^3(256)=t^2(1024)=t(12800)=1$. This observation gives rise to the following questions:

Question 1. Does $t$ have any fixed points? In other words, is there any exp-perfect number?
Question 2. Given a natural number $k\geq 2$, is there a sequence of exp-sociable numbers of length $k$?
Question 3. Is the exponential version of Catalan-Dickson conjecture true? Precisely, is it true that for any natural number $n>1$, the sequence $\{t^{(k)}(n)\}_{k\in \omega}$ ends up becoming constant on $1$ or an exp-perfect number or falling into a loop of exp-sociable numbers?
Note that assuming a negative answer to the questions 1 and 2 (which sounds likely to me), the question 3 will have the following neat formulation: $\forall n>1~~\exists k\geq 1~~~t^{(k)}(n)=1$ which shows a greater degree of nice behavior in comparison with the case of classical aliquot sequences.

 A: Let $p^a$ and $q^b$ be distinct prime powers exactly dividing $n$. (The case of $n$ being a prime power I leave to you.). Then any exp-divisors of $n$ which occur as an exponent must divide both $a$ and $b$, and all the exp-divisors which are bases are powers of some number $c$.
I see I forgot that $n^1$ is not part of the product, so let me throw it in and give an expression for $n$ times $t(n)$.  Let $c$ be the base and $e^j$ the exponent where both $c$ and $e$ are not perfect powers.  Then $$nt(n) = c^{s_1(e^j)}e^{js_0(e^j)/2}.$$
This is because $c$ is raised to all the powers dividing the exponent $e^j$, and the divisors of $e^j$ can be paired off unless $j$ is even. Of course, the formula for $t(n)$ is derivable from this but is not as nice looking.
As a result, this sequence is likely to go to one quickly as the conditions for $t(n)$ to be a perfect power are rare. If e and c share prime factors, then it is especially hard to predict when the result is a perfect power.  I expressed the exponent as a power because that is one of the complications present in analyzing $t(n)$.
Edit 2018.07.21 GRP
So in the formula above, I not only count $n$, I count possible duplicates of divisors which are both a base and an exponent. The revised formula has a fudge factor
$$c^dn t(n) = c^{s_1(e^j)}e^{js_0(e^j)/2}$$
where $d$ accounts for the duplicate powers of $c$ and for which I do not have an expression.
However, this formula shows that in any cycle, the numbers must have all the same prime factors.  Given the tweaks happening to the exponents, I am doubtful of a fixed point, much less a cycle.
End Edit 2018.07.21 GRP
Gerhard "Is Perfect In Many Ways" Paseman, 2018.07.20.
A: Let me add an observation which provides a partial negative answer to the question 1. 
As there are deficient and abundant numbers of the form $2^{2^n}$ with respect to $t$ (e.g. $t(2^{2^2})=2^3<2^{2^2}$ and $t(2^{2^3})=2^{10}>2^{2^3}$), one may hope that $t$ has a fixed point of this form, namely $\exists n~~~t(2^{2^{n}})=2^{2^{n}}$. The following fact states that it is not the case.
Fact. $t$ has no fixed point of the form $2^{2^{n}}$.
Proof. Observe that we have:
$(2^{2^{0}})^{2^n}=(2^{2^{1}})^{2^{n-1}}=(2^{2^2})^{2^{n-2}}=\cdots=(2^{2^{n}})^{2^{n-n}}$
This representation gives us all proper exp-divisors of $2^{2^n}$ that are $2^n, 2^{n-1}, \cdots, 2^1$ and $2^{2^{n-1}}, 2^{2^{n-2}}, \cdots, 2^{2^{0}}$. However, these two lists may have some intersection for some $n$. If we exclude the duplicates and consider the multiplication of all these proper exp-divisors of $2^{2^n}$ we get $t(2^{2^{n}})=2^{\frac{(n+1)n}{2}+2^n - 2^{[\log^{n}_{2}]+1}}$. Consequently $t(2^{2^{n}})=2^{2^{n}}$ if and only if $(n+1)n=2^{[\log^{n}_{2}]+2}$ which is impossible. QED.
Similarly, one may get $t(p^{p^{n}})=p^{\frac{(n+1)n}{2}+\frac{(p^{n}-p^{[\log^{n}_{p}]+1})}{p-1}}$ for prime number $p$ and conclude that $t$ has no fixed point of the form $p^{p^{n}}$ as well.
