Taylor Series Proof for $(1−x+x^2)e^x$

Question

I came across the following problem and have no idea where to start. Here it is:

Prove that every nonzero coefficient of $(1−x+x^2)\text{e}^x$ is a rational number with a numerator of either 1 or a prime.

How does one solve this?

Answer 1

I would guess that the correct question is asking that the numerator of the coefficient of every power greater than $1$ of $x$ in the Taylor series is $1$ or a prime. For $n \gt 1$ the coefficient of $x^n$ of our Taylor series is $$\frac 1{n!}-\frac 1{(n-1)!}+\frac 1{(n-2)!}=\frac {1-n+n(n-1)}{n!}=\frac{n^2-2n+1}{n!}=\frac {(n-1)^2}{n!}$$ If $n-1$ is prime, one factor will cancel with the $n-1$ in the denominator, leaving the numerator $n-1$, which is prime. If $n-1$ is composite and not a square, factor it as $ab$. Then $a,b,n-1$ are different numbers less than $n$ and all will cancel, leaving the numerator $1$. If $n-1$ is a square $ p^2$ greater than $4$, the term is $\frac {p^4}{n!}$ and the terms $p,2p,p^2$ are all different numbers less than $n$ and the numerator will be $1$. Now we check up to $n=5$, find the numerators are all $1$ or prime and we are done.