I want to determine the smallest integer $m$ such that the polynomial $P_{n}(x)$, $n\geq m$, given by : $$\left \lbrace \begin{array}{l} P_{n+1}(x) = P_n(x) (x-n-1) + \prod\limits_{i = 0}^n x-i\\ P_0(x) = 1 \end{array} \right.$$ is not solvable by radicals.
Let $E$ be the set of all roots of $P_n(x)$ and $\mathcal{G} = Gal(\mathbb{Q}(E)/\mathbb{Q})$. I know that $P_n(x) = (-1)^n P_n(n-x)$ and : $$\forall \alpha \in E, 0<\alpha<n.$$
So I deduced :
1- $$\lvert \mathcal{G}\rvert = \prod\limits_{k = 0}^{\lfloor n/2\rfloor -1} n -i -2k,$$ with $n\equiv i\ mod(2)$. So $8\leq m$.
2- $\exists \sigma \in \mathcal{G}, \forall \alpha_j \in E \rvert \sigma(\alpha_j) = \alpha_{n-j-1}$, with $0\leq j < n$. I ordered the roots such that $\alpha_j < \alpha_{j+1}$.
Is it correct (I am new in Galois Theory)? Can we say more about $\mathcal{G}$?