$12^7+8^8$ 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 

You are quite incorrect in your calculations. $8^2 \equiv -1 \pmod {13}$ in your second line. 
However, there is an easier way you can proceed. $$12^{7}+8^8 \equiv (-1)^7+2^{24} \equiv -1+(2^{12})^2 \equiv -1+1 \pmod {13}$$ since $2^{12} \equiv 1 \pmod{13}$ from Fermat's Little Theorem. 
