Modular exponentiation problem

$10^7 \pmod {77}$

I tried repeated squaring, which worked but took many computations. I also tried Fermat's little theorem, but since $7 < 77$ I didn't know how to use it.

Any simpler way to do this?

Answer

Not much effort saved, but work separately mod $11$ (easy, since $10\equiv -1\pmod{11}$) and mod $7$ (easy since by Fermat $10^7\equiv 10\pmod 7$). Then stitch the answers together using the Chinese Remainder Theorem. In this case that can be done by inspection: the answer is $10$.