How to solve $x \equiv 2^{2999999} \pmod {17}$? What trick can I use to solve the question (in the title)?

I need to solve it by hand, wihtout the help of a computer or calculator. And the trick shouldn't just work with my example but also with $3^{123456789}$ for example.
Write down the first few numbers. $2^0 \mod 17 = 1$. $2^1 \mod 17 = 2$. $2^2 \mod 17 = 4$. $2^3 \mod 17 = 8$. $2^4 \mod 17 = 16$. $2^5 \mod 17 = 15$. $2^6 \mod 17 = 13$. $2^7 \mod 17 = 9$. $2^8 \mod 17 = 1$. 

Now if we go on with this, then the results will repeat. We will find that $2^8 = 2^{16} = 2^{24} ... \mod 17 = 1$. Every time the power is a multiple of eight, the modulo is 1. Therefore $2^{2999992} \mod 17 = 1$. And 7 steps further, $2^{2999999} \mod 17 = 9$.
$2^{16} \equiv 1\mod17$,
and $2999999 = 187499 \times 16 + 15$,
so we know $(2^{16})^{187499} \equiv 1\mod17$,
so the answer is $2^{15}+1 \equiv 10\mod 17$.