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.

Building upon OP's idea:

I tried multiplying it by $n+1$ or $n-1$ and then tried factorising it

$$
\begin{align}
(n-1)(n^5+n^4+1) = \color{red}{n^6} \color{blue}{- n^4 + n} \color{red}{- 1} &= \color{red}{(n^3-1)(n^3+1)} \color{blue}{-n(n^3-1)} \\
 &= (n^3-1)(n^3-n+1) \\
 &= (n-1)(n^2+n+1)(n^3-n+1)
\end{align}
$$
