Find the remainder of $1234^{5678}\bmod 13$ Find the reminder of $1234^{5678}\bmod 13$

I have tried to use Euler's Theorem as well as the special case of it - Fermat's little theorem. But neither of them got me anywhere. 
Is there something important here that I am missing. 

$\rm mod\ 13\!:\ \ 12\: \equiv\: -1,\ 10\:\equiv\:-3\ $ so $\rm\ 1234\  =\ 12\cdot 10^{\:2} +\: 34\ \equiv\:\: -1\cdot (-3)^2 + 34\ \equiv\: 25\ \equiv\: -1\:.\ $   
Therefore $\rm\quad 1234^{2\:N}\ \equiv\ (-1)^{2\:N}\ \equiv\ ((-1)^2)^N\ \equiv\ 1^N\ \equiv 1\pmod{ 13}$
