Let $n$ be a positive integer greater than $1$, and define the polynomial $$p_{n}(x)=\sum_{k=0}^{n}\dfrac{x^k}{k!}$$

Is $p_{n}(x)$ irreducible in $\mathbf{Q}[x]$?

I can show it when $n$ is a prime number, since $$n!p_{n}(x)=x^n+nx^{n-1}+n(n-1)x^{n-2}+\cdots+n!x+n!$$ using Eisenstein's criterion.

3more comments