Compute the remainder of $2^{2^{17}} + 1$. This question comes from Fraleighs book A First Course in Abstract Algebra.
Compute the remainder of   $2^{2^{17}} + 1$ divided by 19. [Hint: You will need to compute the remainder of $2^{17} \mod 18$.]
First off I do not even understand how the hint will helps me. The question comes from the section Fermat's and Euler's theorem's.  By fermat's theorem they mean his little theorem: If $a\in \Bbb Z $ and p is prime and p not dividing a, then p divides $a^{p-1} -1$ , that is, $a^{p-1}\equiv 1\pmod p $.
I understand how to do $2^{n}\pmod {19} $. Since 19 does not divide 2, we may say $2^{n-1}\equiv 1\pmod {19} $. Since $2^{n}=2\cdot2^{n-1}\equiv 2\cdot1\pmod {19}$ so in general any $n\in \Bbb Z $ has remainder 2. What is really confusing me is the $+1$ in the statement $2^{2^{17}} + 1$ 

Checking directly, $\,2^9=-1\pmod {19}\,$ , so $\,2^{18}=1\pmod {19}\,$ . Now the hint must be clear:
$$2^4=-2\pmod {18}\Longrightarrow 2^{17}=(2^4)^4\cdot =(-2)^4\cdot 2=16\cdot 2=14\pmod {18} ...$$
