Product of exponents of prime factorization Let $p(n)$ be the product of the exponents of the prime factorization of $n$. For example,
$$p(5184) = p(2^6 3^4) = 24 \;,$$
$$p(65536) = p(2^{16}) = 16 \;.$$
Define $P(n)$ as the number of iterations of $p(\;)$ to reduce $n$ to $1$. For example,
$P(5184) = 3$ because 
$$p(5184)=24, \;p(24) = p(2^3 3^1) = 3, \;p(3)=1 \;;$$
and $P(65536)=4$ because
$$p(65536) = 16, \;p(16)=p(2^4)=4, \;p(4)=p(2^2)=2, \; p(2)=1 \;.$$
Finally, define $m(k)$ to be the minimum value of $n$ such that $P(n) = k$.

Q1. What is $m(k)$?

(I ask this question out of curiosity, not because it is part of a research program.
It was previously posed on MSE.)
It is easy to see that $m(1)=2$, $m(2)=4$, and $m(3)=16$, the latter because $16=2^{2^2}$. But, thanks to Calvin Lin's insight, $m(4)$ is not a power of $2$, but instead is $m(4)=1296= 2^4 3^4$. I do not know the value of $m(5)$.

Q2. More specifically: What is $m(5)$?

I do know that $m(5) > 2 \times 10^8$.
Update.
Will Jagy showed that almost certainly
$m(5) = 2^9 3^6 5^4 7^3 11^2 =9681819840000 \approx 10^{13}$.
As it seems that an explicit expression for $m(k)$ is not in the offing,
I will accept his resolution of Q2 and leave Q1 open.
 A: I neglected to use a special feature of your function, which works the same as the number of divisors function but not the sum of divisors function. That is, if some number $n$ fails to have non-increasing exponents, then there is a smaller number $m,$ constructed by putting the exponents in decreasing order and attaching them to the primes 2,3, etc., such that $p(m)$ is exactly the same as $p(n).$
THEOREM: $m(k)$ has non-increasing exponents in its prime factorization. 
COROLLARY: $m(5) = 2^9 3^6 5^4 7^3 11^2.$
A: Alright, an upper bound on $m(6)$ is
$$ B = 2^{11} 3^{11} 5^7 7^7 11^7 13^5 17^5 19^5 23^5 29^3 31^3 37^3 41^3 43^3 47^3 53^2 59^2 61^2 67^2 71^2 73^2 79^2 83^2 89^2 $$
which is sort of large, granted. 
Next, find the first primorial (product of the consecutive primes beginning with 2) that exceeds $\sqrt B.$ Call the largest prime factor of that primorial $Q = p_r,$ where $p_1=2, p_2=3,$ and so on. I imagine $r < 100$ and maybe $r < 50.$
Finally, run the $r$-tuple loop with nonincreasing exponents on the primes $2,3,\ldots,p_r$ such that each resulting number $N$ given by that prime factorization is less than $B \cdot e^{10}$ by using logarithms, that is $\log N < \log B + 10.0.$
For each such $N$ that satisfies $p(p(p(p(p(p(N)))))) = 1$ but $p(p(p(p(p(N))))) \neq 1,$ print out a line beginning with $\log N$ followed by the $r$-tuple of exponents. Sort. Alternatively, print out nothing, but save $\log N$ and its $r$-tuple in a datatype of some kind, and keep replacing every time a smaller $\log N$ appears. In the end, print out that information. Or, as a sort of hybrid that I like, print out every time $\log N$ decreases. In the beginning, improvements come thick and fast, then slow down as you near the winner. Nice to see some progress reports, you see.
I claim this can actually be done, successfully. Good exercise for certain types of programming class, although the math part may need explaining. Getting a 50 variable multiple loop with certain bounds built in is likely a bit of work...
FRIDAY: At least I was able to find the bound on primes. The big number $B \approx 1.2 \cdot 10^{113},$ and $\sqrt B \approx 3.4 \cdot 10^{56}.$ It suffices to use the first 35 primes in the multiple loop, as the "primorial"
$$ P_{35} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots 137 \cdot 139 \cdot 149 \approx 1.5 \cdot 10^{57}  $$ is larger than $\sqrt B.$
Oh, I always use logs base $e \approx 2.718287828,$ and  $$ \log B \approx 260.37   $$
Given that the loop has 35 variables rather than 100, I can probably do this myself, but certain parts do need to be rewritten in GMP. I also suspect that running time need not be huge, although such predictions sometimes disappear in the face of reality. 
