Compute the remainder of $2^{2^{17}}+1$ when divided by $19$
Hint given in book: Computer the remainder of $2^{17}$ modulo $18$
My attempt:
From Fermat's little theorem, $2^{18}=1(mod19)$
I have done sums where I express the number given as this $LHS ^{something} \times something$ but I can't seem to be able to do that here!
I don't understand how to relate this to what I have to find or how to use the hint given. Please help!
First as I commented $2^{17}=14(mod18)$ i.e $2^{17}=14+18k$ for some $k\in \Bbb{N}$ then $$2^{2^{17}}=2^{14+18k}=2^{14}\cdot2^{{18}k}$$ and since $2^{18}=1(mod19)$ thus$2^{18k}=1(mod19)$ therefore $$2^{2^{17}}=2^{14}(mod19)$$ Now can you end from here?