Calculate $20^{1234567} \mod 251$ I need to calculate the following
$$20^{1234567} \mod 251$$
I am struggling with that because $251$ is a prime number, so I can't simplify anything and I don't have a clue how to go on. Moreover how do I  figure out the period of $[20]_{251}$? Any suggestions, remarks, nudges in the right direction are very appreciated.

If you do not know the little theorem, a painful but -I think- still plausible method is to observe that $2^{10} = 1024 \equiv 20$, and $10^3 = 1000 \equiv -4$. Then, we may proceed like this:
$20^{1234567} = 2^{1234567}10^{1234567} = 2^{123456\times 10 + 7}10^{411522\times 3 + 1} = 1280\times 1024^{123456}1000^{411522} \equiv 1280\times 20^{123456}4^{411522}$.
Observe that after one pass, we are still left with the powers of $2$ and the powers of $20$ that we can handle with the same equivalences above.
We still have to make some calculations obviously (divisions etc), but it is at least not as hopeless as before.
