Proving that a polynomial of the form $(x-a_1)\cdots(x-a_n) + 1$ is irreducible over $\mathbb{Q}$

I want to prove that for any set of distinct integers $a_1,\ldots,a_n$, the polynomial $$h = (x-a_1)\cdots(x-a_n) + 1$$ is irreducible over the field $\mathbb{Q}$, except for the following special cases which are reducible:

$$\left.\begin{cases} a_1 = a\\ a_2 = a+2 \end{cases}\right\} \implies h = (x-a-1)^2$$

and $$\left.\begin{cases} a_1 = a\\ a_2 = a+1\\ a_3 = a+2\\ a_4 = a+3 \end{cases}\right\} \implies h = ((x-a-1)(x-a-2)-1)^2$$

If $h(x)$ is reducible over $\mathbb{Q}$, then by Gauss's Lemma, $h(x)$ is reducible over $\mathbb{Z}$. We can find two monic $p(x), q(x) \in \mathbb{Z}[x]$, both with $\deg(p), \deg(q) < n$, such that $h(x) = p(x)q(x)$.

Notice for $1 \le k \le n$,

$$ p(a_k)q(a_k) = h(a_k) = 1 \land p(a_k), q(a_k) \in \mathbb{Z} \quad\implies\quad p(a_k) = q(a_k). $$

This implies $p(x)$ and $q(x)$ coincide on more points than their degree and hence they are equal to each other, i.e.,

$$\prod_{k=1}^{n} (x - a_k) = h(x) - 1 = p(x)^2 - 1 = (p(x)-1)(p(x)+1).$$

A consequence of this is $n = 2\ell$ is even. Furthermore, relabel $a_k$ is required, we can assume

$$ p(x) - 1 = \prod\limits_{k=1}^\ell (x-a_k) \quad\text{ and }\quad p(x) + 1 = \prod\limits_{k=1}^\ell (x-a_{k+\ell}). $$

If $h(x)$ is reducible over $\mathbb{Q}$, so does $h(x + a)$ for any constant $a$. Using a suitable choice of $a$, we only need to study the special case where one of the $a_k$, say $a_0 = 0$. Under this assumption, we have

$$p(0) - 1 = (-1)^\ell\prod_{k=1}^\ell a_k = 0 \quad\implies\quad p(0) + 1 = (-1)^\ell\prod_{k=1}^\ell a_{k+\ell} = 2. $$ Since $2$ is a prime, there aren't too much choice for $a_{k+\ell}$, they can only be $\pm 1$ or $\pm 2$. The are only $4$ possibilities and only $3$ of them leads to sensible solution. $$\require{cancel} \begin{array}{|r:l|} \hline p(x) + 1 & p(x)-1\\ \hline (x-2)(x-1) & x(x-3)\\ \cancel{(x-2)(x-1)(x+1)} & \cancel{x(x^2-x-1)}\\ (x+2) & x\\ (x+2)(x+1) & x(x+3)\\ \hline \end{array} $$ Putting the offset $a$ back, this leads to following $3$ possibilities for $h(x)$:

$$\begin{align} (x-a)(x-a-1)(x-a-2)(x-a-3) + 1 &= ((x-a)(x-a-3) + 1)^2\\ (x-a)(x-a+2) + 1 &= (x-a+1)^2\\ \cancel{(x-a)(x-a+1)(x-a+2)(x-a+3) + 1} &= \cancel{((x-a)(x-a+3) + 1)^2} \end{align} $$ The $3^{rd}$ set of possibility is not a new one. It can derived from the $1^{st}$ possibility by substitution $a \mapsto a - 3$. This leaves us with two possibilities and it is easy to see they are equivalent to the two exceptions mentioned in question.