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

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 $101!$. 
