Let $F=\mathbb{Z}_p(t)$ for some $p$ prime number. Let $E$ be the splitting field of $f(x)=x^p-t$ over $F$.
(a) We need to show that degree of $E$ over $F$ is $p$, i.e. $[E:F]=p$. (b) Also we need to show that $|G|=1$ where $G=Gal_F(E)$.
My work
Since $f'(x)=px^{p-1}=0$ then $f$ is inseparable, so $[E:F] \leq p$ and so if $[E:F]=d<p$ then there should be a minimal irreducible polynomial of degree $d$ but then this polynomial should divides $f$ which cannot happen as $f$ is inseparable, so $[E:F]$ must be $p$. Is that correct and sufficient argument for (a)?
$(a)$ If $\alpha$ is a root of $f(x)=0$ in $E$ then, $f(x)=x^p-\alpha^p=(x-\alpha)^p$ in $E[x]$ as the characteristic of the field is $p.$ Therefore this shows that, $f$ has only root $\alpha.$ Also, $f$ is irreducible over $F$ as $t\notin F^p.$ So, $E=F[x]/(f(x))$ and hence $[E:F]=p$