Is $1+x+\frac{x^2}2+\dots+\frac{x^n}{n!}$ irreducible?

The polynomial $f(x)=1+x+\frac{x^2}2+\dots+\frac{x^n}{n!}$ often appears in algebra textbooks as an illustration for using derivative to test for multiple roots.

Recently, I stumbled upon Example 2.1.6 in Prasolov's book Polynomials (Springer, 2004), where it is shown that this polynomial is irreducible using Eisenstein's criterion and Bertrand's postulate. However, I do not think the argument is correct. Below you can find the argument presented in the book -- I do not see how Eisenstein is applicable here, since we do not know $p\mid n$. And if we are using Eisenstein's criterion directly to the polynomial $n!f(x)$, this is one of the coefficients that would have to be divisible by $p$. (However, the argument works at least if $n$ is prime.)

So my main question is about the irreducibility of the original polynomial, but I also wonder whether Prasolov's proof can be corrected somehow. To summarize:

• Is the polynomial $f(x)=1+x+\frac{x^2}2+\dots+\frac{x^n}{n!}$ irreducible over $\mathbb Q$?
• Is the Prasolov's proof correct or can it be easily corrected? (Did I miss something there?)

Here is the (whole) Example 2.1.6 from Prasolov's book. The same example is given in прасолов: многочлены(Prasolov: Mnogochleny; 2001,MCCME).

Example 2.1.6. For any positive integer $n$, the polynomial $$f(x)=1+x+\frac{x^2}2+\dots+\frac{x^n}{n!}$$ is irreducible.

Proof: We have to prove that the polynomial $$n!f(x)=x^n+nx^{n-1}+n(n-1)x^{n-2}+\dots+n!$$ is irreducible over $\mathbb Z$. To this end, it suffices to find the prime $p$ such that $n!$ is divisible by $p$ but is not divisible by $p^2$, i.e., $p \le n &lt 2p$.

Let $n = 2m$ or $n = 2m + 1$. Bertrand's postulate states that there exists a prime p such that $m &lt p \le 2m$.

For $n = 2m$, the inequalities $p \le n &lt 2p$ are obvious. For $n = 2m + 1$, we obtain the inequalities $p \le n-1$ and $n-1 &lt 2p$. But in this case the number $n-1$ is even, and hence the inequality $n-1 &lt 2p$ implies $n &lt 2p$. It is also clear that $p \le n - 1 &lt n$. $\hspace{20pt}\square$

Solution 1:

"since we do not know $p∣n$."

I will take Keith's word for it that the argument, overall, is bogus.

However, please note that the condition Prasolov identifies as desired is

$$ p | n! $$ but not $$ p^2 | n! $$

If $p \leq n,$ then $$ n! = 1 \cdot 2 \cdot \cdots (p-1) \cdot p \cdot (p+1) \cdots n$$ so indeed $p | n!$

If $2p \leq n,$ then $$ n! = 1 \cdot 2 \cdots (p-1) \cdot p \cdot (p+1) \cdots (2p-1) \cdot (2p) \cdot (2p+1) \cdots n$$ so $p^2 | n!$

If $p \leq n < 2 p$ then $n!$ is divisible by $p$ but not by $p^2$

EDIT: it is clear from comments that Martin was considering the entire Eisenstein argument, meaning that it is not enough to know the behavior of the last coefficient $n!.$ I did not have the Prasolov book in front of me and was late for an appointment, so I reacted strictly to the excerpt I saw on this site.