Compute $2^{2^{17}}+1$ mod $19$. (hint: compute first $2^{17}$ mod $18$)
Using the fact that there are $6$ numbers coprime to $18$, I got that
$$2^{17}\; \mathrm{mod}\;\; 18 = 2^6 \cdot 2^6 \cdot 2^5 \equiv 2^5 \equiv 14$$ I don't understand, however, how this would help me to compute the original element. Any help would be appreciated.
On
The main point to remember here is that for any element $g$ in a group $G$, we have $g^a = g^b$ if and only if $a$ and $b$ are congruent modulo the order of $g$ in $G$. Here, $G$ is the multiplicative group of integers modulo $19$, and $g = 2$ has order $18$ in this group (this is why the hint tells you to first work modulo $18$, not $19$).
Let me rephrase what you did. You used Euler's Theorem to determine
$$2^{17}=2^{2\times6+5}\equiv2^5\pmod{18}$$
Can you do something similar with $2^{2^{17}}\pmod{19}$?