Sum of digits of sum of digits of powers of 12345 Sum of digits of $12345$ is $1+2+3+4+5=15$. The sum of digits of sum of digits is $1+5=6$.
I have plotted the sum of digits of powers of $12345$ with blue dots (x-axis is the power).
As the average digit is $4.5$, we can see the roughly linear rise with slope $log_{10}(12345) * 4.5 ≈ 18.4$
The red dots represent the sum of digits of sum of digits. Any hint why they are multiplies of $9$? (except the first red dot with value of $6$)

 A: As $12345$ is a multiple of $3$, the powers of $12345$, excluding itself, will be multiples of $9$. By the divisibility rule of $9$ the sum of the digits, and the sum of the sum of the digits will also be divisible by $9$. (That is why the first dot alone was an exception).
A: The sum of digits is divisible by 9 if and only if the number itself is divisible by 9. As 123456 is a multiple of 3 its square and all higher powers are divisible by 9. So the sum of digits, sum of sum of digits, etc. This is a standard test for divisibility by 9.
The reason this test works is as follows. Each digit $a_n$ of a number represents $10^n\times a_n$, where $n=0$ for the units, $n=1$ for the tens, etc. But each power $10^n=1+99...9$ ($(n$ nines). As $99...9$ is divisible by $9$ for any number of $9$s, the number itself is the sum of its digits plus a multiple of $9$. It follows that if the sum of digits is divisible by $9$, so is the number.
A: The powers of $12345$ after $12345$ are all multiples of $9$, because $12345$ is a multiple of $3$, so its powers are multiples of $3^2 = 9$. By the divisibility test for $9$, the sum of digits will also be a multiple of $9$. You can see that on the LHS vertical axis this number is plotted. Even though it looks like it covers every number , it actually covers only multiples of $9$, but seems to do this uniformly enough.
After that, the red lines are the sum of sum of digits. The point to realize is that even if a number is very large its sum of digits could be really small. For example, $10000000000001$ has sum of digits $2$ and so on.
So the RHS vertical line represents this quantity, but it is going to be really small. How?
Let us derive a simple bound for the sum of digits. If $n$ is a $k$ digit number, then the maximum sum of digits is $9k$. But $n \geq 10^{k}$ so $\log_{10}n \geq k$. Finally, the desired bound is $9 \log_{10} n$.
Finally, if $n = a^b$ for some $a,b$ then $\log_{10} n = b \log_{10} a$ so the bound becomes $9b \log_{10} a$.
The bound of "sum of sum of digits" would then be :
$$
9 \log_{10} [9 b \log_{10} a]
$$
which even in $b$ is logarithmic. That is why even when $b$ is like $20000$ or $40000$ and $a$ is of the same order, the sum of sum of digits is not going to be very large, in fact it is going to be $\log \log$ order smaller then the number.
You can imagine then what "sum of sum of sum of digits" would be like. 
