Show $1+x+(x^2/2!)+ \cdots + (x^n/n!)=0$ has no rational solutions for all $n>1$. 
Prove that the equation $$1+x+\frac{x^2}{2!}+ \cdots + \frac{x^n}{n!}=0$$ has no rational solutions for all $n>1$.

Assume there is a rational solution $\frac{p}{q} \in \mathbb{Q}$ with $(p,q)=1$, then by clearing denominator we have $p\mid n!$ and $q= \pm 1$. Hence the solution must be an integer solution. Furthermore, by considering it modulo $2$, $p$ must be even. But then I cannot proceed anymore.
Please help. Thanks in advance.
 A: For $|q|\ne 1$ Zubin's argument is fine. Otherwise, I have got an idea which I am presenting below. Please point out if something seems wrong.
For $|q|=1$, the solution $x$ is a negative integer , then you can show that the equation yields $$n!\left(1+x^2/2!+\cdots+x^{2k}/(2k)!\right)=n!\left(x+x^3/3!+\cdots+x^{2k+1}/(2k+1)!\right)$$ where $k=[n/2]$. Then $x|n!\Rightarrow n!=ax$ for some positive integer $a$. Then, we can see that this yields $a|x\Rightarrow x=ab$ which yields $n!=a^2b$ and in a similar manner we can see that this will result in $b|a$. So basically we can generate in this way a sequence $a_1,a_2,\cdots$ such that $a_1|x,a_2|a_1,\cdots$ and $n!=a_1x=a_1^2a_2=a_2^3a_3$ This will continue until we get some $m>1$ such that $a_m=1$ and then we will get $n!=a_{m-1}^m$. I think this is not possible which will lead us to a contradiction. I am trying to prove this last statement.
A: Suppose, for contradiction, that there exist $p,q,n \in \mathbb{Z}$ such that $n>1$ and $q>0$ and $\gcd(p,q)=1$ and $$1+\frac{p}{q}+\frac{1}{2!}\left(\frac{p}{q}\right)^2+\frac{1}{3!}\left(\frac{p}{q}\right)^3+\cdots+\frac{1}{n!}\left(\frac{p}{q}\right)^n =0$$
Multiplying by $n! \cdot q^n$, we get
$$q^nn!+pq^{n-1}n!+\frac{n!}{2!}p^2q^{n-2} + \cdots + \frac{n!}{n!}p^n=0$$
If we take the equation modulo $q$ we get $$p^n \equiv 0 \pmod{q}$$
This means $q \mid p^n$, but this is impossible since $\gcd(p,q)=1$.

Edit: The above answer is not correct. It only shows that $q=1$ and $p<0$.
