Showing a polynomial is irreducible in $\mathbb{Q}[x]$ The Problem

Let $a_0=1$ and recursively define $a_n$ to be the smallest prime number strictly greater than $a_0+...+a_{n-1}$
  then $(a_n)_{n\geq 0 }$, it follows that $$a_0x^n+a_1x^{n-1}+...+a_n$$ is irreducible in $\mathbb{Q}[x]$

Another way of defining $a_n=\min\{p|p>a_0+...+a_{n-1} \text{ and } p \text{ is prime}\}$
Here is the first couple numbers in the sequence $(a_n)_{n\geq 0}$: 
$(1,2,5,11,23,43,89,...)$
Here is the first couple of polynomials:
$$1$$
$$x+2$$
$$x^2+2x+5$$
$$x^3+2x^2+5x+11$$
$$x^4+2x^3+5x^2+11x+23$$
$$x^5+2x^4+5x^3+11x^2+23x+43$$
$$x^6+2x^5+5x^4+11x^3+23x^2+43x+89$$
My Thoughts
It is clear that if we can find some prime $p$ where $q_n(x)=a_0x^n+a_1x^{n-1}+...+a_n$ is irreducible in $\mathbb{Z_p}[x]$ then $q_n(x)$ is irreducible in $\mathbb{Q}[x]$. With that said I think that I have to prove:

$q_n(x)=a_0x^n+...+a_n$ is irreducible in $Z_{a_{n-1}}[x]$

but I am kind of stuck here. Any Ideas?
 A: Gauss' lemma states that if a polynomial is irreducible over integers, it is also irreducible over rationals. Let's show the polynomial 
$$P_n(x)=a_0x^n+a_1x^{n−1}+...+a_n$$
such that $$a_n > a_{n-1}+a_{n-2}+...+a_0 \tag{1}$$ 
$a_k \in \mathbb{N}^{*},\forall k$ and $a_n$-prime is irreducible over integers.

Proposition 1 If $\alpha \in \mathbb{C}$ is a root of $P_n(x)$, i.e. $P_n(\alpha)=0$, then $|\alpha|>1$.
Let's suppose $|\alpha| \leq 1$ then 
$$a_0\alpha^n+a_1\alpha^{n−1}+...+a_n=0 \Rightarrow |a_n|=|a_0\alpha^n+a_1\alpha^{n−1}+...+a_{n-1}\alpha| \Rightarrow \\
a_n\leq a_0|\alpha^n|+a_1|\alpha^{n−1}|+...+a_{n-1}|\alpha| \leq a_0+a_1+...+a_{n-1}$$
which contradicts $(1)$

Proposition 2 $P_n(x)$ is irreducible over integers.
Let's suppose it's reducible, i.e. $P_n(x)=G(x)\cdot F(x)$, where $G, F$ are non-constant polynomials with integer coefficients. Then $P_n(0)=G(0)\cdot F(0)=a_n$ which is prime. This means that the absolute values of the last coefficient of $G$ is 1 and of the last coefficient of $F$ is $a_n$ (or vice-versa, but WLOG let's assume so). Let's notes
$$G(x)=b_kx^k+...+b_0, b_i\in \mathbb{Z}, b_k \ne 0, |b_0|=1$$
From Vieta's formulas we have that the absolute value of the product of $G(x)$'s roots ($G(\gamma_i)=0, \gamma_i \in \mathbb{C}$) is 
$$\left|\prod_{i} \gamma_i \right|= \left|\frac{b_0}{b_k}\right|\leq 1$$
But every single root of $G(x)$ is also a root for $P_n(x)$ and, according to Proposition 1, we must have
$$\left|\prod_{i} \gamma_i \right|=\prod_{i} \left|\gamma_i \right|>1$$
So we have a contradiction.

Here is a recommended source of inspiration.
