The trend of huge number (4)

$\text{Let}\; 2017^{2017^{2017}}=n.\text{Then:}\\ 1^n+2^n+3^n+\cdots +2016^n\pmod{2017}\\ (1^n+2016^n)+(2^n+2015^n)+(3^n+2014^n)+\cdots+(1008^n+1009^n)\pmod{2017}\\ (\color{#3D99F6}{1}^n+\color{#3D99F6}{(-1)}^n)+(\color{#3D99F6}{2}^n+\color{#3D99F6}{(-2)}^n)+(\color{#3D99F6}{3}^n+\color{#3D99F6}{(-3)}^n)+\cdots+(\color{#3D99F6}{1008}^n+\color{#3D99F6}{(-1008)}^n)\pmod{2017}\\ n\;\text{is odd,so:}\\ (-a)^n=-a^n\implies \boxed{a^n+(-a)^n=0}\\ \implies \sum^{1008}_{j=1} (\color{teal}{j^n+(-j)^n})=\sum^{1008}_{j=1} \color{#D61F06}{0}=\color{#20A900}{\boxed{\boxed{0}}}$

Relevant wiki: Fermat's Little Theorem

$\large 1^{{2017}^{{2017}^{{2017}}}}+2^{{2017}^{{2017}^{2017}}}+3^{{2017}^{{2017}^{2017}}}+\dots+2016^{{2017}^{{2017}^{2017}}} \equiv x \pmod{2017}$

$\large 1+2+3+\dots+2016 \equiv x \pmod{2017}$ (By Fermat's little theorem)

$\large \frac{(2016)(2016+1)}{2} \equiv x \pmod{2017}$

$\large (1008)(2017) \equiv x \pmod{2017}$

$\large x \equiv 0 \pmod{2017}$