Assume true for x=k ie assume k^5-k=10c (where c is an integer)
Then look at
(k+1)^5 - (k+1)
Expand out the binomial, the ones will cancel.
Replace the terms k^5-k with 10c.
At this stage you shpuld have:
10c + 5k^4 +10k^3 +10k^2 +5k
=10(c + k^3 + k^2) + 5 (k^4 + k)
The first part of this is obviously a multiple of 10.
The last bit is obviously a multiple of 5 but also k^4+k is always even
(if k is odd, k^4 is odd and so k^4+k is even; if k is even, k^4 is even and so k^4+k is even), so the last bit is a multiple of 5 and 2 ie a multiple of 10.......