# Show that the field of fractions of $\mathbb{Z}[[x]]$ is properly contained in $\Bbb Q((x))$

I've been working on this for a while but I don't know how to proceed. Here it's what I've done: Clearly, my goal is to show that every element of the field of fractions is in $\Bbb Q((x))$ and to show that there exist an element of $\Bbb Q((x))$ that is not an element of the field of fractions. The problem suggests considering the series expansion for $e^x$ so that's what I did. I guess that that's the hint because $e^x$ is the element in $\mathbb{Q}((x))$ that is not in $\text{Frac}(\mathbb{Z}[[x]])$ I'm looking for. So $$e^x=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$$ Since $n!\in\mathbb{N}$ then $\frac{1}{n!}\in \mathbb{Q}$ and $e^x\in \mathbb{Q}((x))$ Now, this is where I can't go further. I've tried assuming that $e^x\in \text{Frac}(\mathbb{Z}[[x]])$ and using that it must be a unit to achieve a contradiction but I don't know how to do so. Any hints? Thanks in advance

Write $e^x = A/B$ with $A,B \in {\mathbb Z}[[x]]$. We may assume that the constant terms of $A,B$ are not zero. Then write $B = b_0 + b_1 x + \cdots$. Now take $R := {\mathbb Z}[ b_0^{-1} ]$ and $S := R[[x]]$. Then $B$ is a unit in $S$. So we can write $A/B = C$ for some $C \in S$. But we can't write $e^x$ as an element of $S$.
Suppose $e^x\cdot q(x)=r(x)$, with $q$ and $r$ in $\mathbb{Z}[[x]]$. If $p$ is a prime, then the coefficient of $x^p$ in $e^x\cdot q(x)$ is $$\sum_{i=0}^p \frac{q_i}{(p-i)!}$$ where $q_i$ is the coefficient of $x^i$ in $q(x)$. Since this has to equal $r_p$, we can mutiply both sides by $p!$ to get $$q_0 + p(\cdots)=p!r_p$$ so that $p$ divides $q_0$. Since $p$ was an arbitrary prime, we see that all primes divide $q_0$, and so $q_0=0$. We can then proceed inductively to show $q(x)=0$, and thus $r(x)=0$ as well.