Mental remainder

$\phi(2017) = 2016$ where $\phi$ is the Euler's Totient function.
$a^{\phi(n)+1} \equiv a \pmod{n}$ $S = \displaystyle \sum_{k=1}^{4035} k^{2017} \equiv \sum_{k=1}^{4035} k \equiv \dfrac{4035\cdot4036}{2} \equiv 4035 \cdot 2018 \equiv 1 \pmod{2017}$

Modular arithmetic rules:

$a_1 \equiv b_1 \pmod n$ and $a_2 \equiv b_2 \pmod n$ then 1. $a_1 + a_2 \equiv b_1 + b_2 \pmod n\$ and 2. $a_1 a_2 \equiv b_1 b_2 \pmod n$

Now note that $a^{2017}+(4034-a)^{2017}\equiv 0 \pmod {2017}\$ (when a is a positive natural number less than 2017). To see this take $a \equiv b \pmod {2017}\$ and by 2. we have $a^{2017} \equiv b^{2017} \pmod {2017}\$ now $(4034-a) \equiv -b \pmod {2017}\$ as is readily seen by noting that 2017 divides 4034 and again by 2. we get that $(4034-a)^{2017} \equiv -b^{2017} \pmod {2017}\$ and by 1. the result follows. Now by pairing the summands as described above we easily see that [and the middle term $2017^{2017}$ is left alone for which (mod 2017) is clearly zero] $\sum_{n=1}^{4034} n^{2017}\equiv 0 \pmod {2017}\$ and as is easily seen by applying 2. $4035^{2017}\equiv 1 \pmod {2017}\$ so it follows that $\sum_{n=1}^{4035} n^{2017}\equiv 1 \pmod {2017}\$ And as the name suggests this can be done mentally with little focus :). I would love to see alternative solutions!