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).

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(I'm not sure about the relevant tags so please feel free to alter them!) –  Dave May 19 '13 at 15:41
That is solved by $92093$, just hope that we have at least one user on this who have solved it with a proof. –  Inceptio May 19 '13 at 15:53
@Inceptio I expect the vast majority of those users used Java's built-in BigInteger to do this without any thought, but I consider that cheating. –  Dave May 19 '13 at 15:54
You are supposed to use $\mod 9$, since it uses the digit sum. –  Inceptio May 19 '13 at 15:57
@Inceptio won't that give the repeated sum? This is just looking for the sum. –  Dave May 19 '13 at 15:59

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.