Truncated expresion of $e^x$ is an irreducible polynomial I need to prove that the truncated expression of the Taylor development of the exponential function is irreducible.
$ 1 + \frac{1}{1!}x + \frac{1}{2!}x^2+... + \frac{1}{p!}x^p \in \mathbb{Q}[x]$
where p is prime. Searching for answers I have found some proofs that use Galois theory, but that is very complex for me to understand. I know about Eisenstein's and reduction criterion.
Eisenstein's criterion
Suppose we have the following polynomial with integer coefficients.
$
{\displaystyle Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}
$
$
{\displaystyle Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}
$
If there exists a prime number p such that the following three conditions all apply:

*

*$p$ divides each $a_i$ for $0 ≤ i < n$,

*$p$ does not divide $a_n$, and

*$p^2$ does not divide $a_0$,

Reduction criterion
(converse of reduction criterion?)
Any help would be appreciated. Thank you :).

Let us denote $$P(x)=1 + \frac{1}{1!}x + \frac{1}{2!}x^2+ \dots + \frac{1}{p!}x^p.$$ Then we have that
$$p!P(x)= p!+ p!x + \frac{p!}{2!}x^2+ \dots + p(p-1)x^{p-2}+px^{p-1}+x^p$$ satisfies the Eisenstein's criterion, from where it follows that $p!P(x)$ is irreducible. And thus $P(x)$ is irreducible.
