$N^7 - N = N(N^6 - 1)$
If N = 6k then the remainder will be 0 because 6|N.
If N = 6k +- 1, then $N^6 = (6k)^6 \pm something*(6k)^5 + .... \pm something*(6k) + 1$ so $N^6 - 1 = 6(a\ bunch\ of\ stuff)$ so the remainder will be 0 because $6|N^2 - 1$
If N = 6k +/- 2, the $N^6 = (6k)^6 \pm something*(6k)^5 + .... \pm something*(6k) + 2^6$ so $N^6 - 1 = 6(a\ bunch\ of\ stuff) + 63$. So $N^7 - N = N(N^6 - 1) = (6k \pm 2)(6(a\ bunch\ of\ stuff) + 63) = 6(\ even \ more \ stuff) \pm 126 =6(\ even \ more \ stuff) \pm 21*6 $ so the remainder is 0 because $6|N^7 - N$.
== late edit == Just realized the obvious
$M = N^7- N = N(N^3 - 1)(N^3 + 1)= N(N^2 + N + 1)(N - 1)(N^2 - N + 1)(N + 1)=(N-1)N(N+1)(N^2 - N + 1)(N^2 + N + 1)$
Which may or may not make things easier. Let's see.
If $N$ is even then 2 divides $N$. If $N$ is odd then 2 divides $N+1$. So 2 divides $M$.
If 3 divides $N$ then 3 divides $M$. If $N = 3k \pm 1$ then 3 divides $N \mp 1$. So 3 divides $M$.
So $6|M$. In fact 6 divides $N^3 - N$ while $N^7 - N$ = $(N^3 - N)(N^4 + N^2 + 1)$.
In fact, 6 divides the product of any three consecutive integers and $N^3 - N$ can be written as a product of three consecutive integers.
In hindsight, I'm a little surprise how weak this question is. To start with it's a multiple choice question so if there is one constant remainder it has to be 0 as N = 6k would have 0 remainder. Second it's a very easy and well, known fact that any 3 consecutive integers are divisible by 6. (I'm kicking myself for not realizing that earlier.) And there so much more to $N^7 - N$ then the division by 6. There's there entire factor $N^4 +N^2 + 1$ which is completely superfluous.
So here's a related not much more difficult puzzle, Prove $N^5 - 5N^3 + 4N$ is divisible by 120.