Since 42=2*3*7, the puzzle is equivalent to proving that n^7-n is divisible by 2,3, and 7.

Divisibility by 7:

If p is a prime and a is any integer not divisible by p, then a^(p-1) -1 is divisible by p. (Fermat)

Now, n^7-n = n(n^6-1), and 7 is prime, so unless 7 divides n, (n^6-1) is divisible by 7. Note that when n is 7, or a multiple of 7, the initial factor n guarantees divisibility by 7.

Divisibility by 2,3: 

Note the factorization: n^6-1 = (n^3-1)(n^3+1), while (n-1)(n^2+n+1) = (n^3-1), and (n+1)(n^2-n+1) = (n^3+1).

So n^7-n has factors (n-1),n,(n+1) three consecutive numbers of which at least one must be a multiple of 2 and at least one a multiple of 3.

As was to be shown.