(n+1)^7 =  n^7 + 7 n^6 + 21 n^5 + 35 n^4 + 35 n^3 + 21 n^2 + 7 n + 1

subtract (n+1):

n^7 + 7 n^6 + 21 n^5 + 35 n^4 + 35 n^3 + 21 n^2 + 6 n  

 

The difference between this and n^7 - n is

7 n^6 + 21 n^5 + 35 n^4 + 35 n^3 + 21 n^2 + 7 n 

This factors into

7 n (n + 1) (n^2 + n + 1)^2  (thanks to Wolfram Alpha)

7 n (n + 1) is always a multiple of 14. If n is a multiple of 3 or one less than a multiple of 3, it is also so, as n(n+1) will be a multiple of 3. And if n is one more than a multiple of 3, (n^2 + n + 1) will be a multiple of 3.

So the difference from one n to the next will always be a multiple of 14*3=42, so if any n^7-n is a multiple of 42, then n+1 will also be so. If n=1, n^7-n = 0, which is a multiple of 42, so adding a multiple of 42, as in the preceding paragraph gives another, etc.