Question: Suppose today is Wednesday. What day of the week will it be $10^{{10}^{10}}$ days from now?
Answer: This is what I have tried so far. I know that you need to find $10^{{10}^{10}}\text{mod} 7$ to solve this problem. If $10^{{10}^{10}} \text{mod}7=0$, then it will be Wednesday, if $10^{{10}^{10}}\text{mod}7=1$, it will be Thursday, and so on. I tried breaking up the large exponent as follows: $10^{{10}^{10}}\text{mod}7=(10^{{10}^5} \text{mod}7\cdot 10^{{10}^5}\text{mod7})\text{mod}7$. However, I do not know how to find $10^{{10}^5} \text{mod}7$. The answer to this problem is Sunday, so that means that $10^{{10}^{10}}\text{mod}7=4$.
Note that $10^{6}\equiv 1\pmod{7}$. Then, to solve this problem we would like to find $10^{10}\pmod{6}$. This is equivalent to $(-2)^{10}\equiv 1024\equiv 4\pmod 6$. Thus, $10^{10^{10}}\equiv 10^{6k + 4}\pmod{7}$ for some integer $k$. Thus, $10^{10^{10}}\equiv 10000\equiv 4\pmod{7}$. Can you take it from here?