prove equation has no rational root.

Question

Prove

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For all $n>1$, equation $\sum _{k=1}^n \frac{x^k}{k!}+1=0$ has no rational root.

I'm not sure whether there are two questions,for without brace after Sigma.

My thought is to prove it is not reducible on rational field.

Answer 1

Multiplying by $n!$ we can make all coefficients to become integers. By the rational root theorem, the root has to be integer and divides $n!.$ Moreover, since all coefficients except for the first one are divisible by $n,$ $x^n$ is divisible by $n.$ Take any prime divisor $p$ of $n.$ Then $p$ divides $x$ and it is enough to show that $x^k\frac{n!}{k!}$ is divisible by the higher power of $p$ than $n!$ for each $1\le k\le n$ The power of $p$ that $k!$ is divisible is less than $k/p+k/p^2+...=\frac{k}{p-1}\le k$ provided $p\ge 2$ and the result follows.