We completely factor:







To be divisible by 30, the above must be divisible by 2, 3, and 5

3 consecutive integers n-1, n, and n+1 are always such that one of
them is divisible by 2 and another is divisible by 3. thus n5-n is always
divisible by 6.

So we only need to show that one of the four factors is divisible by
5.

For any integer n, every sequence of n consecutive integers can be written
as nk+a,nk+a+1,nk+a+2,...,nk+a+n-1, for some integers k and a.

We can let n=5, a=-3

Then every sequence of 5 integers can be written 5k-3, 5k-2, 5k-1, 5k, 5k+1 

n is divisible by 5 if n=5k
n-1 is divisible by 5 if n=5k+1 because (5k+1)-1 = 5k 
n+1 is divisible by 5 if n=5k-1 because (5k-1)+1 = 5k
n²+1 is divisible by 5 if n=5k-2 because (5k-2)²+1 = 25k²+10k+4+1 =
      25k²+10k+5 = 5(5k²+2k+1)
n²+1 is divisible by 5 whenever n=5k-3 because (5k-3)²+1 = 25k²-30k+9+1 =
      25k²+15k+10 = 5(5k²+3k+2)

So for every sequence of 5 consecutive integers, one of the 4 factors of
n5-n is divisible by 2,3,5, which means n5-n is always
divisible by 30.

[PROVED]

Edwin