The remains

$\displaystyle \sum_{i=1}^{2016} i^{2016} \equiv 201 \cdot (\sum_{i=1}^9 i^{2016}) + \sum_{i=1}^6 i^{2016} \equiv 201(33) + 5 \equiv 8 \pmod{10}$ , because $11 \equiv 21 \equiv ... \equiv 1 \pmod{10}$ ,,, $22 \equiv 12 \equiv 2 \equiv ... \equiv 32 \pmod{10}$ ...

$1^{2016} \equiv 1 \pmod{10}$

$2^{2016} \equiv (2^4)^{504} \equiv 6 \pmod{10}$

$3^{2016} \equiv (3^2)^{1008} \equiv (-1)^{1008} \equiv 1 \pmod{10}$

$4^{2016} \equiv 6 \pmod{10}$

$5^{2016} \equiv 5 \pmod{10}$

$6^{2016} \equiv 6 \pmod{10}$

$7^{2016} \equiv 1 \pmod{10}$

$8^{2016} \equiv 6 \pmod{10}$

$9^{2016} \equiv 1 \pmod{10}$

we do not consider all numbers like 10,20,30,...2010 raise to the power 2016 as they will not affect the remainder when divided by 10 as they are going to give zero remainder. we need to consider the rest of the numbers for our answer.If we carefully observe the pattern we see that there are 201 sets of $\sum _{ i=1 }^{ 9 }{ { i }^{ 2016 } }$ and last 6 terms in the numerator. calculating $\sum _{ i=1 }^{ 9 }{ { i }^{ 2016 } }$ =33 using cyclicity and remaining last six term add up to 25 taking only the last digit of each term for calculation , for our answer we have multiply 201X33=3(last digit) and add 5(last digit of 25) and divide 5+3=8 by 10 we get 8 as our answer.