Proving that $\gcd(2016, 2017^{2017} + 2018^{2018}) = 1$ 
Prove that $\gcd(2016, 2017^{2017} + 2018^{2018}) = 1$.

I don't know how to do this. I tried different strategies, they all leading me nowhere. I tried to write $2017^{2017} + 2018^{2018}$ as $2017 x + 2017y$, where $x, y$ are some integers, but I don't know how. 
Also, tried proving by contradiction but I think that's the wrong direction too.
Any help/hints?
EDIT: I'm suppose to find this without using a calculator. So I don't know the prime factorization etc.
Based on Arthur's hint, I have $$ (2017^{2017} + 2018^{2018}) \mod 2016 \\ = (1 + 2^{2018} \mod 2016) \mod 2016. $$ I now still need to prove that $$2^{2018} \mod 2016 = 0. $$ 
 A: The primes dividing $2016$ are $2,3$ and $7$. So to solve our problem we must show neither of these primes divides $2017^{2017}+2018^{2018}$.
Clearly $2017^{2017}+2018^{2018}$ is odd.
Notice $2017^{2017}+2018^{2018}\equiv 1^{2017}+(-1)^{2018}\equiv 2 \bmod 3$
Finally, notice $2017^{2017}+ 2018^{2018}\equiv 1^{2017}+2^{2018}\bmod 7$.
By Fermat's theorem $2^{6}\equiv 1 \bmod 7$, so $1^{2017}+2^{2018}\equiv 1+(2^{336})^6\times 2^2\equiv 1+2^2\equiv 5 \bmod 7$
A: Clearly, $ 2017^{2017} + 2018^{2018}$ is odd and the highest power of $2$ in $2016$ is $5, 2016=2^5\cdot63$
So, it sufficient to show $(63,2017^{2017} + 2018^{2018})=1$
As $2017\equiv1\pmod{2016},2017^{2017}\equiv1\pmod{2016},2017^{2017}\equiv1\pmod{63}$
Again, $2018\equiv2\pmod{63}, 2018^{2018}\equiv2^{2018}$
Now $2^6\equiv1\pmod{63}$ and $2018\equiv2\pmod6\implies2^{2018}\equiv2^2\pmod{63}$
$\implies2017^{2017} + 2018^{2018}\equiv1+2^2\pmod{63}$
Can you take it from here?
A: ${\rm mod}\,\ 2016\!:\ 2^{11}\!=2048\equiv 2^{\large 5}\! \overset{\,\times\, {\large 2^{\Large 6}}}\Rightarrow 2^{\large 17}\!\equiv 2^{\large 11}\!\equiv 2^{\large 5}\ldots\Rightarrow \color{#c00}{2^{\large 5+6K}\!\equiv 2^{\large 5}}\, $ for all $K\ge 0\:$ so 
$$ \color{#0a0}{2^{\large 2018}}\equiv\, 2^{\large 3}\, \color{#c00}{2^{\large 5 + 6\cdot 335}}\equiv 2^{\large 3}\,\color{#c00}{2^{\large 5}}\equiv \color{#0a0}{2^{\large 8}}$$
$$\Rightarrow\ \  N :=\, 2017^{\large 2017}\!+2018^{\large 2018}\equiv\, 1^{\large 2017}\!+\color{#0a0}{2^{\large 2018}}\equiv 1+\color{#0a0}{2^{\large 8}}\equiv\, \color{#90f}{257}\qquad $$
Thus Euclidean algorithm $\,\Rightarrow (2016,N) = (2016,\, N\, {\rm mod}\ 2016) = (2016, \color{#90f}{257}) = 1$
A: Hmmph.  The two things that jump at me are:
1) $2017^{2017} = (2016 + 1)^{2017} \equiv 1 \mod 2016$
$2018^{2018} = (2016 + 2)^{2018} \equiv 2^{2018} \mod 2016$
And
2) $2017^{2017} + 2018^{2018}$ is odd. So $2 \not \mid \gcd(2016, 2017^{2017}+2018^{2018} )$.
Therefore
$\gcd(2016=63*2^5, 2017^{2017}+2018^{2018}) = \gcd(63, 1 + 2^{2018})$
Which should be solvable by FLT
By FLT $2^{\phi(63)} = 2^{\phi(3^2)\phi(7)} = 2^{3*2*6}= 2^{36} \equiv 1 \mod 63$
So $2018 \equiv 2 \mod 36$ so $2^{2018}\equiv 2^2 = 4 \mod 63$.
So $1 + 2^{2018} \equiv 4 = 1 = 5 \mod 63$.
So $\gcd(2016, 2017^{2017} + 2018^{2018})= \gcd(63, 5) = 1$.
