I'm not looking for a full solution, only a hint please!
Let $\zeta$ be a $p$-th root of unity in an algebraic closure of $Q_p$. Show that $Q_p(\zeta) = Q_p ((-p)^{\frac{1}{p-1}})$.
Following a hint in the exercise, I write $p = (1- \zeta)^{p-1} \varepsilon$ for some $\varepsilon \in Q_p(\zeta)$ and try to show that $-\varepsilon$ is $p-1$th potence in $Q_p(\zeta)$. For this it would be sufficient to see that
a) $-\varepsilon \in R'$ (the valuation Ring of $Q_p(\zeta)$)
b) $-\varepsilon \equiv 1 \mod p$
For a) I can simply compute the absolute value of $(1- \zeta)^{p-1}$ and deduce that $|\varepsilon|=1$. But how can I see that $\varepsilon \equiv -1 \mod p$ and not some other unit?
I've tried some arithmetic tricks with $p = (1- \zeta)^{p-1} \varepsilon$ to see that $\varepsilon^2 \equiv -\varepsilon \mod p$, but it was fruitless.
For b) you can use Wilson's theorem, and the finite geometric series, i.e., $$ \epsilon=\frac{p}{(1-\zeta)^{p-1}}=\prod_{i=1}^{p-1}\frac{1-\zeta^i}{1-\zeta}=\prod_{i=1}^{p-1}\sum_{j=0}^{i-1}\zeta^j \equiv (p-1)!\equiv -1 \mod (1-\zeta) $$