Transcendental numbers in $\mathbb{Q}_p$ I'm trying to prove that $K((x))/K(x)$ is not an algebraic extension, where $K$ is a field of characteristic $p>0$, and $K((x))$ is the field of fractions of $K[[x]]:=\{\sum_{i=0}^{\infty}a_ix^i\mid a_i\in K\}$.
To make it more concrete, I'm considering the case $K=\mathbb{Z}/p\mathbb{Z}$ and $x=p$, so that we need to find $\tau\in\mathbb{Q}_p$ transcendental in respect to the field extension $\mathbb{Q}_p|\mathbb{Q}$. 
I thought about $\tau:=\sum_{i=0}^{\infty}\frac{p^n}{n!}$, which I've checked converges in the $p$-adic metric, but I'm not sure is transcendental or not. It's very tempting to say that $e^p=\sum_{i=0}^{\infty}\frac{p^n}{n!}$, and since $e$ is transcendental, we're done, but I know this probably doesn't make any formal sense. How do I resolve this?
 A: My only suggestion would be to try to imitate the ordinary proof of Liouville’s Theorem, which can be used to show that $\sum_n10^{n!}$ is transcendental over $\Bbb Q$. This would mean showing that $\sum_n t^{n!}$ is transcendental over $\kappa(t)$, for (say) $\kappa$ a finite field.
A: Maybe showing that a certain element is transcendental is not easy. I think that about the same or more work is required as to prove that $e^p\in\mathbb{R}$ is transcendental. 
One way to get around is to use the cardinality comparison. This way is not constructive, but we can prove that there is $\tau\in\mathbb{Q}_p$ which is transcendental over $\mathbb{Q}$. 
Since $\mathbb{Q}$ is countable, the set of algebraic elements inside $\mathbb{Q}_p$ over $\mathbb{Q}$ is countable. However, the set $\mathbb{Q}_p$ is uncountable. Thus, there must be a transcendental element $\tau\in\mathbb{Q}_p$. 
This argument is possible for any countably infinite field $K$.   Set-theoretically stating, the algebraic closure $\overline{K}$ over $K$ satisfies
$$
|\overline{K}|=|K| = \aleph_0 < 2^{\aleph_0} = |K^{\mathbb{N}}|=|K((x))|.
$$
