= Solution
Expand the translated <cyclotomic polynomial>:
$$
\frac{(1+T)^p-1}{T}=p+\binom p2T+\cdots+\binom p{p-1}T^{p-2}+T^{p-1}.
$$
Every nonleading coefficient is divisible by $p$, and the constant term is not divisible by $p^2$. The polynomial is <Eisenstein>, hence irreducible. Its root $\lambda=\zeta_p-1$ is a uniformizer of $E=\mathbb Q_p(\zeta_p)$, and the <Eisenstein polynomial> valuation criterion gives
$$
\boxed{[E:\mathbb Q_p]=e(E/\mathbb Q_p)=p-1.}
$$
Thus this local cyclotomic extension is totally ramified.
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