If n = 1 then n^10+n^5+1 = 3, a prime. For the rest of this solution assume n > 1.

Start with an identity: n^15 - 1 = n^15 - 1.  And we'll factor each side in different ways.
(n^3-1) * (n^12+n^9+n^6+n^3+1) = (n^5-1) * (n^10+n^5+1)

(n-1) * (n^2+n+1) * (n^12+n^9+n^6+n^3+1) = (n-1) * (n^4+n^3+n^2+n+1) * (n^10+n^5+1)

Cancel the n-1 factors:

(n^2+n+1) * (n^12+n^9+n^6+n^3+1) = (n^4+n^3+n^2+n+1) * (n^10+n^5+1)

The factors of n^2+n+1 must be distributed among n^4+n^3+n^2+n+1 and n^10+n^5+1.

At this point we'll show n^2+n+1 and n^4+n^3+n^2+n+1 are coprime.

n^4+n^3+n^2+n+1 = (n^2+1) * (n^2+n+1) - n^2.  So n^2+n+1 and n^4+n^3+n^2+n+1 can only have a common factor if n^2 has a common factor with n^2+n+1.  

But n^2 and n^2+n+1 are coprime, therefore n^2+n+1 and n^4+n^3+n^2+n+1 are coprime.

Then n^2+n+1 must be a factor of n^10+n^5+1.  But when n>1 both expressions evaluate to different integers bigger than 1.  Thus n^10+n^5+1 must be composite for n>1.

Then the only natural number such that n^10+n^5+1 is prime is n=1.