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 $P(n) = k$.

> <b>Q1</b>. 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][1].)

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][2], $m(4)$ is not a power of $2$, but instead is $m(4)=1296= 2^4 3^4$. I do not know the value of $m(5)$.

> <b>Q2</b>. More specifically: What is $m(5)$?

I do know that $m(5) > 2 \times 10^8$.
<hr /><b>Update</b>.
Will Jagy showed that almost certainly
$m(5) = 2^9 3^6 5^4 7^3 11^2 \approx 10^{13}$.
As it seems that an explicit expression for $m(k)$ is not in the offing,
I will accept his resolution of <b>Q2</b> and leave <b>Q1</b> open.

  [1]: http://math.stackexchange.com/q/507909/237
  [2]: http://math.stackexchange.com/a/507914/237