How to prove this polynomial is irreducible over $\mathbb{Q}$ $f(x)＝x^{n}+x+p$ where $p\geq 3$ is a prime.Prove that $f(x)$ is irreducible over $\mathbb{Q}$.
My try:Let $y＝x+m$ for some $m$ and try to use Eisenstein's Criterion.But I can't find the suitable $m$.
Thanks in advance for the help.
 A: .Suppose that $\alpha$ is a complex zero of $f = x^n + x + p$. If $|\alpha| \leq 1$ , then $|p| \leq |\alpha^n + \alpha| \leq 2$, is a contradiction. So $|\alpha| > 1$ is true.
Suppose that $f = gh$, where $g,h$ are non-constant integer polynomials. Note that $p = g(0)h(0)$. Since $p$ is prime, we have either $|g(0)| = 1$ or $|h(0)| = 1$. Say that it is the case with $g$, and let $b$ be the leading coefficient of $g$. 
If $\alpha_i$ are the roots of $g$,  then their product $|\alpha_1 \alpha_2 \ldots \alpha_k| = \frac 1{|b|} \leq 1$, so that at least one of these roots must have magnitude less than or equal to $1$. However, these are roots of $f$ as well, so this cannot happen. Hence, the contradiction gives irreducibility.

Addendum : The same technique can be used to prove that if $f(x) = a_nx^n + ... + a_0$ is a polynomial with integer coefficients such that $|a_0| > |a_1| + ... + |a_n|$, and $|a_0|$ is prime, then $f$ is in fact irreducible over the rationals. So for example, $x^n - 53 x^2 + 27x - 83\ $ is irreducible for the same reason.

