Kronecker-Weber Theorem I'm stuck with an article "A simple proof of Kronecker-Weber Theorem" on this website. On page 7, the author proofs that $\mathbb{Q}_p((-p)^{\frac{1}{p-1}}) = \mathbb{Q}_p(\zeta_p)$. While I understand the reasoning of the proof ($[\mathbb{Q}_p((-p)^{\frac{1}{p-1}}):\mathbb{Q}_p] = [\mathbb{Q}_p(\zeta_p):\mathbb{Q}_p]$, so if one contains the other, we are done), I don't get how he makes his $u$ to use for Hensel's Lemma.
Specifically, he defines a polynomial
$$g(X) = \frac{(X+1)^p-1}{X} = X^{p-1}+pX^{p-2}+\ldots +p,$$
so of course we have $g(\zeta_p-1)=0$, but I don't get how
$$g(\zeta^p-1)\equiv (\zeta_p-1)^{p-1}+p (\bmod(\zeta_p-1)^p).$$
Apart from understanding this lemma, is there an other (simpler) proof of this fact ?
Any help would be appreciated.
 A: I think the point is that this cyclotomic extension is totally ramified, i.e. $p = u(\zeta - 1)^{p - 1}$ for some unit $u$ (we could surely work this out). The inner coefficients of $g(X)$ are all divisible by $p$, so when we multiply by $(\zeta - 1)^i$, $i \geq 1$, then we get something that is divisible by $(\zeta - 1)^p$.
A: Here is a general argument (which I won't quite complete, but which it shouldn't be hard to complete): 
The extension $\mathbb Q_p(\zeta_p)$ is cyclic of degree $p-1$ over $\mathbb Q_p$.
Since $\mathbb Q_p$ contains the $p-1$st roots of $1$, it must be a Kummer extension, i.e. obtained by extracting a $p-1$st root of some element of $\mathbb Q_p$.  It is also a totally ramified extension, so it must be extracting a $p-1$st root of a uniformizer.  It remains to work out what the precise uniformizer is; this shouldn't be hard.  (Of course, the answer is $-p$.)
A: You should recall that the ideal $(p)$ decomposes into $(1-\zeta_p)^{p-1}$, so $(\zeta_p-1)^p = (p (\zeta_p-1))$.
Thus $g(\zeta_p - 1) = (\zeta_p - 1)^{p-1} + p + \sum_{1<n<p} \binom{p}{n} (\zeta_p - 1)^{n-1}$.
Since the binomial coefficient is a multiple of $p$, each term in the sum is in the ideal $(p(\zeta_p-1))$
