# Sum of digits raised to a power

Let $S$ equal the sum of the digits of $2014^{2014}$. Let $T$ equal the sum of the digits of $S$. Let $U$ equal the sum of the digits of $T$.

What is $U$?

Because $10^n\text{ mod }9=1$, the sum of digits $\text{mod }9$ is the number itself $\text{mod }9$. $$2014^{2014}\equiv 7^{2014}\equiv 7\cdot(7^3)^{671}\equiv7\cdot1^{671}\equiv7\quad\text{(mod }9)$$ What remains is to check that $U$ can't be bigger than 7. The least such number would be $16$.
Clearly, $2014^{2014}<10^{4\cdot2014}$, so $S$ is at most $9\cdot8056<10^5$. Similarly $T$ is at most $9\cdot5=45$, so $U$ is at most $4+9=13$. Therefore the answer is $7$.
I don't know what $U$ is, but I can give an upper bound for it. $$2014^{2014} < 3000^{3000} = 27000000000^{1000} = (2.7*10^{10})^{1000} < (10^{11})^{1000} = 10^{11000}$$
So $2014^{2014}$ is at most $11000$ digits long, which means $S$ is at most $99000$, which means $T$ is at most $9+8+9+9+9=44$, which means $U$ is at most $3+9=12$.