Showing that the fraction field of $\Bbb Z[[t]]$ is strictly contained in the Laurent formal series field $\Bbb Q((t))$
A hint is to consider $e^t=1+t+{t^2\over2}+...+{t^n\over n!}+...$
So $e^t\in\Bbb Q((t))$ is obvious. But why isn't it in the fraction field of $\Bbb Z[[t]]$? after all ${t^n\over n!}$ is in Frac$(\Bbb Z[[t]]$) $\forall n\in\Bbb Z$
Assume $e^t\in \operatorname{Frac}(\mathbb{Z}[[t]])$, then there exist $f,g\in \mathbb{Z}[[t]]$ coprime such that $e^t = \frac{f}{g}$. Thus, we get $f=e^t g$. For $n\in \mathbb{N}$ we get
$$ f_n = \sum_{j=0}^n \frac{g_{n-j}}{j!}$$
and thus
$$ g_0 = n!f_n - \sum_{j=0}^{n-1} \frac{n! g_{n-j}}{j!}.$$
Set $n=p$ for some prime and deduce that $p$ divides $g_0$. Hence, we have $g_0=0$, but then we get
$$ f_0= \frac{g_0}{0!} =0 $$
this contradicts the assumption that $f,g$ were coprime (as now $f$ and $g$ have the common factor $t$).