A question about some reasonings in Example 1.3 from these notes on unramified extensions.
Example 1.3. Consider the finite unramified extensions of $\mathbb{Q}_p$. By the above theorem, these are in 1-1 correspondence with finite extensions of $\mathbb{F}_p$. But $\mathbb{F}_p$ has a unique extension of degree $n$ for every $n$, namely the splitting field of $x^{p^n}-x$. It follows that $\mathbb{Q}_p$ has a unique unramified extension of degree $n$ for each $n$, obtained as the splitting field of $x^{p^n}-x$, i.e. by adjoining the $p^n-1$ st roots of unity. Moreover, the maximal unramified extension $\mathbb{Q}_p^{nr}$ of $\mathbb{Q}_p$ corresponds to the separable (=algebraic) closure of $\mathbb{F}_p$, and so is obtained by adjoining the $p^n-1$ st roots of unity for all $n$.
Next part I not understand:
For any integer $n$ with $(n, p):= gcd(n,p) = 1$, we have $p^{\phi(n)}-1 ≡ 0 \mod n$ (note that $\phi(n)$ is the Euler function counting the order of the multiplicative group $(\mathbb{Z}/(n))^*$, so we see that $\mathbb{Q}_p^{nr}$
is obtained by adjoining the $n$ th roots of unity for $(n, p) = 1$ for
all $n$, i.e. as the direct limit of the fields $\mathbb{Q}_p(\zeta_n)$.
The three last lines I not understand. Why $\mathbb{Q}_p^{nr}$ which we in first observed as direct limit of $\mathbb{Q}_p(\zeta_{p^n-1})$ running over all $n \in \mathbb{N}$ can be also expressed as direct limit of $\mathbb{Q}_p(\zeta_n)$ where now $n$ is running over naturals with $(n,p)=1$.
Why is this a direct consequence of fact that if $(n, p):= gcd(n,p) = 1$ then $p^{\phi(n)}-1 ≡ 0 \mod n$? This simply implies that $\mathbb{Q}_p(\zeta_n) \subset \mathbb{Q}_p(\zeta_{p^{\phi(n)}-1})$. Why this implies $\bigcup_{n \in \mathbb{N}} \mathbb{Q}_p(\zeta_{p^n-1}) =\bigcup_{n: \ (p,n)=1} \mathbb{Q}_p(\zeta_n)$? Can the argument presented above be explained in detail? I not understand it.
$gcd(n,p)=1$ implies that (the residue of) $p$ is a unit in the ring $\mathbb Z/n$. The units of any ring $R$ form a group, usually denoted by $R^*$. If $G$ is any finite group with neutral element $1$, and $g \in G$, then it is well-known that $g^{ord(G)} = 1$. By definition, $\phi(n) = ord((\mathbb Z/n)^*)$.
Put that together and we have $p^{\phi(n)} -1 = 0$ (mod $n$).
As you say yourself, this implies that $\mathbb Q_p(\zeta_n) \subseteq \mathbb{Q}_p(\zeta_{p^{\phi(n)}-1})$ and so
$$\bigcup_{n \in \mathbb N: (p,n)=1} \mathbb Q_p(\zeta_n) \subseteq \bigcup_{n \in \mathbb N: (p,n)=1}\mathbb{Q}_p(\zeta_{p^{\phi(n)}-1}) \subseteq \bigcup_{k \in \mathbb N}\mathbb{Q}_p(\zeta_{p^{k}-1})$$.
But conversely, clearly $(p^k-1, p)=1$ for any $k \in \mathbb N$, so
$$\bigcup_{k \in \mathbb N}\mathbb{Q}_p(\zeta_{p^{k}-1}) \subseteq \bigcup_{n \in \mathbb N: (p,n)=1} \mathbb Q_p(\zeta_n) $$
and hence all $\subseteq$ are equalities $=$.