Does there exist a power-summer of order $+ \infty$?

We start with some number $n$ and sum its digits (we can denote sum-of-digits function as $S_d$) to obtain number $S_d(n)$.

If $S_d(n)$ is prime then we calculate number $n^2$ and sum its digits to obtain $S_d(n^2)$. If $S_d(n^2)$ is prime then we calculate number $n^3$ and sum its digits to obtain $S_d(n^3)$, and so on...

We can call number $n$ a power-summer of order m if the numbers $S_d(n),...S_d(n^m)$ are all primes.

We can call number $n$ a power-summer of order $+ \infty$ if $n$ is power-summer of order m for every $m \in \mathbb N$

A question is:

Does there exist a power-summer of order $+ \infty$?

Are you of the opinion that there is some global maximum, that is, a natural number $W$ such that order of every $n$ is less than $W$?

An answer is not in my reach, I do not know much about sum-of-digits functions, but maybe someone has some good ideas.

Peter found a number of order $14$, a number $20619661$ and calculated that upto $n=10^9$ there is no number with an order greater than $14$.

from Hot Weekly Questions - Mathematics Stack Exchange