Irreducibility of a polynomial in a finite field The question asks me to show that $\mathbb{Z}_{13}[x]/\langle x^{2014}-x^{1000}+1 \rangle$ is a field.

Now what I know is that since $\mathbb{Z}_{13}$ is a field, so if I show that $x^{2014}-x^{1000}+1 $ is irreducible, then $\langle x^{2014}-x^{1000}+1 \rangle$ will be a maximal ideal and so $\mathbb{Z}_{13}[x]/\langle x^{2014}-x^{1000}+1 \rangle$ will be a field.

But I don't see how to show the polynomial is irreducible in $\mathbb{Z}_{13}[x]$. For the degree $\le 3$, I could use whether there is a root or not but I don't know for such a big degree. 

Maybe there is a duplicate for this question, but please help anyway. Thank you.
This is not true.  The polynomial $x^{2014}-x^{1000}+1$ has $x^2+6$ as a factor, and in particular is not irreducible.  You can quickly verify this by observing that if $x^2=-6$ then (working in $\mathbb{Z}_{13}$) $$x^{2014}-x^{1000}+1=(-6)^{1007}-(-6)^{500}+1=(-6)^{11}-(-6)^8+1=0.$$
HINT

Use the fact that for $p=13$ prime, $x^{p-1}\equiv 1 \pmod p$.  The resulting t3nth degree polynomial is much easier to work with.