The last digit of $3^{2017}$ is 3, so the last digit of the numerator is 5.
$$(a_1+a_2)+(a_2+a_3)+\dots+(a_{2016}+a_{2017})+(a_{2017}+a_1) = 2(a_1+a_2+\dots+a_{2017}).$$
The sum of these 2017 nonnegative integers being even, one of the summands, and hence their product, is even, so the last digit of the number
$$9^{(a_1+a_2)(a_2+a_3)\dots(a_{2016}+a_{2017})(a_{2017}+a_1)}$$
is 1, and the denominator is divisible by 10.
Hence, the fraction can be simplified by 5.