AIME 2017b Problem 8

Find the number of positive integers $n$ less than $2017$ such that $$1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}$$ is an integer.