TITLE: How to prove that field of rational functions is a *proper* subset of field of formal Laurent series?
QUESTION [8 upvotes]: Now, if $F$ is a field, I can prove easily that $F(x)\subseteq F((x))$ but I'm having problems to show this is a proper inclusion.

If for example $F=\Bbb R$ or $\Bbb C$, I can take a well-known function, say $\cos x$ and use its power series to show proper inclusing, because if 
$$\left(F((x))\ni\right)\;\cos x=1-\frac{x^2}2+\frac{x^4}{24}-\cdots=\frac{f(x)}{g(x)}\in F(x)$$
then, since $f(x),g(x)\in F[x]$ (polynomials), we'd get that $\cos x$ has a finite number of zeros, which is absurd, and thus $\cos x\in F((x))\setminus F(x)$.
My problem now is: what to do if the field $F$ is not one of the usual, infinite ones? For example, if $F$ has positive characteristic?
Any input will be duly appreciated.

REPLY [8 votes]: An easy argument : $1-x$ has a square root in $F((x))$ but not in $F(x)$.
Indeed, in $F((x))$, take the usual Taylor expansion of $\sqrt{1-x}$ at $0$. But in $F(x)$, looking at decomposition in irreducible factors easily shows that such a square root can't exist.
If the characteristic of $F$ is $2$, this won't work, you will have to take a cubic root.
Maybe I can comment a little on the argument : what I'm using is the fact that $F((x))$ is henselian, whereas $F(x)$ is not. This is a quite different argument from the one by @user26857 : they use the fact that some elements of $F((x))$ are not solutions of algebraic equations (over $F(x)$), I use the fact that some elements of $F((x))$ are solutions of such equations. This is really a different argument because there are henselian extensions of $F(x)$ which are algebraic over $F$ (namely the henselianization).