Irreducibility of $\sum_{n=0}^{101}\frac{x^n}{n!}$

Question

What can be said about the irreducibility of $\sum_{n=0}^{101}\frac{x^n}{n!}$ over the field of rationals?

Multiplying and dividing throughout by $101!$, we obtain the given polynomial is irreducible iff the polynomial $\sum_{n=0}^{101}P_{n}^{101}x^n$, where $P_r^n$ is the number of permutations of $r$ items from $n$ items. The new polynomial does not seem to have a root according to rational root theorem. But, how do we proceed further? I hope it is irreducible over $\mathbb{Q}[x]$. Any hints? Thankds beforehand.

Answer 1

According to Schur's theorem all truncated exponentials of form $\sum_{n=0}^{k}\frac{x^n}{n!}$ are irreducible over rationals, see for example Keith Conrad's blurb Irreducibility of Truncated Exponentials for correct proof. In fact the following more general result is proven there using algebraic number theory:

Any polynomial $$1+c_1X+c_2\frac{X^2}{2!}+\dots+c_{n-1}\frac{X^{n-1}}{(n-1)!}\pm \frac{X^n}{n!}$$ with $c_i \in \mathbb{Z}$ is irreducible in $\mathbb{Q}[X]$.

Yours is just a special case, although as pointed out in comments/other answer that case can be proven much simpler.

Answer 2

Thanks to @ThomasAndrews comment, the problem is straightforward, the modified polynomial $\sum_{n=0}^{101}P_{n}^{101}x^n$ clearly satisfies the Eisenstein criterion with the prime number $101$ dividing all coefficients except the last and the square of $101$ not dividing the leading coefficient of $101!$.