Elements of $\mathbb{Q}(\zeta_p)$ fixed under $\zeta_p \mapsto \zeta_p^g$

Question

I'm reading Harold Edwards’ Galois theory and before going into Galois theory he was discussing about Gauss' works on constructible $n$ gons (so please avoid Galois Theory while answering this question).

Fix a prime $p$ and a primitve root $g$ of $p$. Let $\zeta_p$ be a primitive $p$-th root of unity.

Let $S: \mathbb{Q}(\zeta_p) \mapsto \mathbb{Q}(\zeta_p)$ be the automorphism replacing $\zeta_p \mapsto \zeta_p^g$, and let $K_d$ be the set of all elements of $\mathbb{Q}(\zeta_p)$ invariant under $S^d$. Now he claims the following things:

1. It's easy to see that $K_1 = \mathbb{Q}$ and $K_{p-1} = \mathbb{Q}(\zeta_p)$ and if $m | n | p-1$ then $K_m \subset K_n$.
2. It's not hard to see see if $d | D | p-1$ then $K_D$ is a simple algebraic extension of $K_d$
3. Pick a chain $1 = d_0 | d_1 | \cdots | d_z = p-1$ such that every divisor of $p-1$ apperars exactly once. Then consider the tower of fields $K_{d_0} \subset K_{d_1} \subset \cdots K_{d_z}$. Then each element of $K_{d_{i+1}}$ can be expressed as terms of radicals with elements of $K_d$ via methods of Lagrange resolvents.
4. Let $\omega$ be a primitive $p-1$th root of unity. Then if $\sum_{0 \leq i \leq p-2} P_i(\omega) \zeta_p^i = 0$, then $P_i(\omega) = 0$ for all $i$ and this is "hard".

OK, so I don't understand all of them properly and need help understanding them.

1. OK, this is pretty clear.
2. I don't see it why $K_D$ should be a simple algebraic extension of $K_d$ ? The best I can get is this far: Define $\displaystyle \tau_d := \sum_{0 \leq i < \frac{p-1}{d}-1} \zeta_p^{g^{di}}$. Then $\tau_d \in K_d$ but it's not in any $K_m$ for which $m | d$ and $m < d$. But then I'm not sure how $\tau_D$ can be expressed as a root of a polynomial in $\mathbb{Q}(\tau_d)$.
3. Why this should be true ?
4. Isn't this very obvious (I don't get why Gauss found it very hard) ? So we are basically wanting that $[\mathbb{Q}(\omega, \zeta_p): \mathbb{Q}(\omega)] \geq p-1$, but doesn't this follows from if you let $\phi$ to be $p(p-1)$th root of unity then $\mathbb{Q}(\phi) = \mathbb{Q}(\omega, \zeta_p)$, and thus $[\mathbb{Q}(\omega, \zeta_p): \mathbb{Q}(\omega)] = \frac{[\mathbb{Q}(\omega, \zeta_p): \mathbb{Q}]}{[\mathbb{Q}(\omega): \mathbb{Q}]} = \frac{[\mathbb{Q}(\phi): \mathbb{Q}]}{[\mathbb{Q}(\omega): \mathbb{Q}]} = \frac{\phi(p-1)(p-1)}{\phi(p-1)} = p-1$ ?