# Irreducible polynomial $p_{n}(x)=\sum_{k=0}^{n}\frac{x^k}{k!}$ for all positive integers $n$

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.

• Summation starts from $k=0$, right? May 30 '16 at 6:46
• There is a fairly elementary proof at mattbakerblog.wordpress.com/2014/05/02/… together with some more notes on the topic. May 30 '16 at 8:50
That follows from a theorem of Schur saying that any polynomial $\sum_{k=0}^nc_k\frac{x^k}{k!}$ with $c_i\in\mathbf{Z}$, $c_0,c_n\in\{1,-1\}$, $n\ge 1$, is irreducible over $\mathbf{Q}$.