How can I show that the polynomial $t^n-x$ is irreducible in $F(x)[t]$?

A short way of seeing this:

By Gauss' Lemma, $t^n - x$ is irreducible in $F(x)[t]$ if and only if it is irreducible in $F[x][t]$. But $F[x][t] = F[t][x]$ and $t^n - x$ is a polynomial of degree $1$ in $F[t][x]$ so is irreducible in $F[t][x]$ and therefore in $F[x][t]$ so as well in $F(x)[t]$.

Hint: The Eisenstein criterion can be generalized to arbitrary integral domains and (in the case of UFDs) their fraction field as well.

Suppose $F$ has characteristic $0$ or at least than its characteristic is prime to $n$. Suppose for the moment that $F$ contains $n$ $n^{th}$ roots of unity, and let $\zeta$ be a primitive such root. Consider $F(x^n)$, a subfield of $F(x)$. The automorphism of $F(x)$ taking $x$ to $\zeta x$ has order $n$ and fixed field $F(x^n)$. (This is an endomorphism because you get to map the indeterminate to anything. Then you have to check (easy) that it's onto.) This shows that $t^n-x^n$ is irreducible over $F(x^n)$. But since $F(x)$ is isomorphic to $F(x^n)$ by the map taking $x$ to $x^n$, we also have $t^n-x$ irreducible over $F(x)$.

Since the degree of $k(a,b)$ over $k(a)$ is never larger than the degree of $k(b)$ over $k$, you can readily see that this argument holds even if $F$ does not contain the desired roots of unity, as long as they can be adjoined.

For the case $(n, char F)>1$, you can handle that, too, but it will be a different argument since it's not a separable extension. I leave that as an exercise.