$12^7+8^8$ divided by $13$ I Need to find what the remainder is when $12^7+8^8$ is divided by $13$
I have a solution, but don't know if it is right. 
$12=-1\mod13$
$12^7=-1\mod13$
$8=8\mod13$
$8^2=6\mod13$
$8^4=10\mod13$
$8^8=9\mod13$
Then I did $-1\mod13+9\mod13=8\mod13$, so the remainder is $8$.
If someone could tell me if this is correct, it would be appreciated. Thanks 

By Fermat's Little Theorem we have $2^{12}=1\bmod13$. So $8^4=1\bmod13$ and hence $8^8=1\bmod13$. As you said, $12^7=(-1)^7=-1\bmod13$, so $12^7+8^8=0\bmod13$.
