Prove that $n^5+n^4+1$ is composite for $n>1.$ 
Prove that $f(n)=n^5+n^4+1$ is composite for $n>1, n\in\mathbb{N}$.

This problem appeared on a local mathematics competition, however it looks like there is no simple method to solve it. I tried multiplying it by $n+1$ or $n-1$ and then tried factorizing it, but it was too tough for me.
Any help will be appreciated.

$$n^5+n^4+1 = n^5 - n^2 + n^4 + n^2 + 1 = n^2(n^3-1) + (n^2+1)^2 - n^2 = n^2(n-1)(n^2+n+1) + (n^2-n+1)(n^2+n+1) = (n^3-n+1)(n^2+n+1)$$
