HINT for summing digits of a large power I recently started working through the Project Euler challenges, but I've got stuck on #16 (http://projecteuler.net/problem=16)

$2^{15} = 32768$ and the sum of its digits is $3 + 2 + 7 + 6 + 8 = 26$. What is the sum of the digits of the number $2^{1000}$?

(since I'm a big fan of generality, my interpretation is to find a solution to the sum of digits of $a^b$ in base $c$, and obviously I'm trying to solve it without resorting to "cheats" like arbitrary-precision numbers).
I guess this is simpler than I'm making it, but I've got no interest in being told the answer so I haven't been able to do a lot of internet searching (too many places just give these things away). So I'd appreciate a hint in the right direction.
I know that $2^{1000} = 2^{2*2*2*5*5*5} = (((((2^2)^2)^2)^5)^5)^5$, and that the repeated sum of digits of powers of 2 follows the pattern $2, 4, 8, 7, 5, 1$, and that the last digit can be determined by an efficient pow-mod algorithm (which I already have from an earlier challenge), but I haven't been able to get further than that… (and I'm not even sure that those are relevant).
 A: In this case, I'm afraid you just have to go ahead and calculate $2^{1000}$. There are various clever ways to do this, but for numbers this small a simple algorithm is fast enough.
The very simplest is to work in base 10 from the start. Faster is to work in binary, and convert to base 10 at the end, but this conversion is not completely trivial. A compromise (on a 32-bit machine) would be to work in base 1000000000, so that you can double a number without overflow. Then at the end the conversion to base 10 is much simpler.
A: Just note that no problem in Project Euler requires you to use bigger numbers than $2^{64}-1$ (or $2^{63}-1$ for signed). However sometimes it is just too painful to avoid big numbers, so it is just easier to use them. 
That being said, I don't think there is a mathematical trick for this problem, although I think the problem really just asks you to implement your own multiplication in terms of base $10$ digits. You know already how to do that on paper! Try $2\cdot 2$, $4 \cdot 2$, $8 \cdot 2$, $\dots$ by hand, notice how digits are carried over. Try to formalize this into an algorithm that uses array for digits. Test it for small values, then use it for $2^{1000}$ (just repeat $1000$ times). Finally it is only about the sum.
It might feel a bit unsatisfactory that all you are required to do is multiplication, but hey, it is just $16$-th problem yet!
A: Spoiler alert
It's not the most honorable solution, but in C# you can use the BigInteger class....
 var powerDigits = BigInteger.Pow(2, 1000).ToString();
 var sum = 0;
 for (var i = 0; i < powerDigits.Length; i++)
     sum += (int) Char.GetNumericValue(powerDigits[i]);
 Console.WriteLine(sum); //1366

A: Here are 2 hints that really helped me along to get to the solution:
1)

"We can use an array, or hash map, to store the digits of any big
number. Every time we multiply each position (the values) of the
array/hash map by two, we need to start from the ‘One’ position to
carry over the digits.
With this mechanism, we can multiply 1000 times 2 quite quickly. We
don’t even need 1000 arrays – one array is enough when we sum the
result directly."

Source: https://www.educative.io/edpresso/what-is-the-power-digit-sum
You may want to stop reading that page before looking at the implementation code towards the bottom of that page.
2.

"Consider an array of ints with the digits, and repeatedly multiply by
2."

Source: https://www.hackerrank.com/contests/projecteuler/challenges/euler016/forum/comments/189057
This one isn't extremely insightful, but it did help light a lightbulb in my head. To go a bit further (and without revealing everything), how can you cache results as you "repeatedly multiply by 2"?
Edit: Note that my comment #2 is a rewording of Sil's answer (https://math.stackexchange.com/a/2879132/162331) above, particularly "Try 2⋅2, 4⋅2, 8⋅2, … by hand, notice how digits are carried over. Try to formalize this into an algorithm that uses array for digits.".
