Proving that the map $k^*/N_{K/k}(K^*) \to \operatorname{Br}(K/k)$ is injective. I'm trying to understand the proof of the following result in number theory:

Let $K/k$ be a cyclic extension and let $\sigma$ be a generator of its Galois group. The map $a\mapsto [(\sigma,a)]$ induces an isomorphism of groups
$$k^*/N_{K/k}(K^*) \to \operatorname{Br}(K/k),$$
where $(\sigma,a)$ is the cyclic algebra generated by $\sigma$ and $a\in k^*$.

The part where I'm stuck is in proving that the kernel of the map above is contained in $N_{K/k}(K^*)$. That is, we want to prove that if $(\sigma,a)$ is split, then $a$ is the norm of an element in $K^*$.
I'll try my best to translate the proof of this fact that is present in the notes I'm reading:

Let us show that the induced map $k^*/N_{K/k}(K^*) \to \operatorname{Br}(K/k)$ is injective. Let $a\in k^*$ and suppose that $A=(\sigma,a)$ is split. Then $A\cong (\sigma,1)$. Fix an embedding of $K$ in $A$ and an element $\alpha\in A$ such that $\alpha^n=1$ and $x\alpha=\alpha\sigma(x)$ for every $x\in K$.

Everything is ok so far. The proof then proceeds as:

By Noether-Skolem, we can find $\gamma\in A$ and $\beta\in A$ such that $\beta^n=1$ and $x\alpha=\alpha\sigma(x)$ for every $x\in \gamma K\gamma^{-1}.$

This seems like a typo to me... But let's go on with the proof.

Then $(\gamma^{-1}\beta\gamma)\alpha^{-1}$ commutes with $K$ and so it is an element $x$ of $K$. (I'm ok with this statement.) We conclude that
$$1=\alpha^n=\beta^{-n}N_{K/k}(x)=N_{K/k}(x).$$

I don't know how $(\gamma^{-1}\beta\gamma)\alpha^{-1}$ commutes with $K$ nor why $\beta^{-n}N_{K/k}(x)$ is equal to $\alpha^n$. My calculations seem to show that $\alpha^n N_{K/k}(x)=\gamma^{-1}\beta^n \gamma$. The latter is not a real problem, since it indeed implies the result but the former is.
I want to understand how we can use Noether-Skolem to find elements $\gamma$ and $\beta$ such that $(\gamma^{-1}\beta\gamma)\alpha^{-1}$ commutes with $K$.
 A: We can prove that $\alpha$ lies in the norm group $N_{K/k}(K^{*}) $ in the following way. I will use without proof, the result that a crossed product algebra $[L,K,\{,\}]$ (where $L/K$ is Galois with Galois group $G$, and $\{,\}$ represents a $2$-cocycle $\{,\}: G*G \rightarrow L^{*}$ satisfying $\{a,b\}\{ab,c\}=\{a,bc\}\phi_a(\{b,c\})$ for all $a,b,c$ in $G$) is a matrix ring $M_n(K)$ if and only if the $2$-cocycle is split, that is there exists a function $f: G \rightarrow L^{*}$ such that $\{a,b\}=f(a)\phi_a(f(b))f(ab)^{-1}$ for all $a,b$ in $G$. (This can be proved using Noether-Skolem and is a special case of the $B(L/K) \rightarrow H^2(G, L^{*})$ isomorphism). Anyway, so our cyclic algebra $[K, k, \sigma, a]$ where say $[K:k]=n$ corresponds by definition to the $2$-cocycle given by
$\{\sigma^r, \sigma^s \}=1$ if $0 \leq r, s, r+s<n$
$\{\sigma^r, \sigma^s \}= \alpha$ if $0 \leq r, s <n \leq r+s$.
So, if $\alpha$ lies in kernel of map $k^{*} \rightarrow B(K/k)$, then the above cocycle is split. Thus there is a map  $f: G \rightarrow L^{*}$ such that $\{a,b\}=f(a)\phi_a(f(b))f(ab)^{-1}$ for  all $a, b$ in $G$, where $G=Gal(K/k)$ is cyclic group. Now, we see that
$\alpha= \prod_{g \in G} \{g, \sigma \}$
(since unless $g= \sigma^{-1}$, we have $\{g, \sigma \}=1$ and $\{ \sigma^{-1}, \sigma \}= \alpha$)
which equals
$\prod_{g \in G} f(g) \phi_g(f(\sigma)) f(g \sigma)^{-1}$
and since $g \rightarrow g \sigma$ is a bijection $G \rightarrow G$, this cancels out to
$\prod_{g \in G} \phi_g(f(\sigma))=N_{K/k} (f( \sigma))$
so $\alpha=N_{K/k}(f(\sigma))$ is a norm. Proved
