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. 



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


I. Schur, Einige Sätze über Primzahlen mit Anwendungen auf Irreduzibilitätsfragen I, Sitzungsberichte
Preuss. Akad. Wiss. Phys.-Math. Klasse 14 (1929), 125–136. Also in Gesammelte Abhandlungen, Band
III, 140–151


See also [here](http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/schurtheorem.pdf).


