Tensor product of Galois extension Let $K/k$ be a finite Galois extension of fields with Galois group $G$. How to show that the (n+1)-fold tensor product 
$$K \otimes_k K \otimes_k K \cdots \otimes_k K$$  is isomorphic to $$\prod\limits_{G^n} K$$ (where the index set $G^n$ is the n-fold product of $G$) by the map  $$x_0 \otimes x_1 \otimes \cdots x_n \mapsto (x_0g_1(x_1)g_1g_2(x_2) \ldots g_1g_2\ldots g_n(x_n))_{(g_1, \ldots, g_n)}$$
It is claimed as part of a proof of something else) in Stephen Shatz’s book Profinite Groups, Arithmetic, Geometry.

 A: This is how I'd do it. With a primitive element $K=k[a_1] \cong k[x_0]/(f(x_0))$ then $${\bigotimes_k}^nK\cong {\bigotimes_k}^nk[x_0]/(f(x_0)) \cong k[X]/I,\qquad X=(x_0,\ldots,x_n),I=(f(x_0),\ldots,f(x_n))$$ 
Let $A=(a_1,\ldots,a_d)\in K^d$ the roots of $f$ and for each function $j:1 \ldots n \to 1 \ldots d$, $A^j=(a_1,a_{j(1)},\ldots,a_{j(n)})$. 
The homomorphisms $k[X]/I \to K$ are given by $X \mapsto g(A^j)$ for each $j$ and $g \in Gal(K/k)$. The image is $K$ so the kernel is a maximal ideal $\mathfrak{m}_j$. 
The $\mathfrak{m}_j$ are different maximal ideal, so they are comaximal, there are $d^n$ of them, so $[k[X]/I:k] = \sum_j [k[X]/\mathfrak{m}_j:k]$, thus we must have $$I = \prod_j \mathfrak{m}_j$$ and $${\bigotimes_k}^n K \cong k[X]/I= k[X]/\prod_j\mathfrak{m}_j\cong\prod_j k[X]/\mathfrak{m}_j \cong \prod_j K$$
Next write $A^j = (a_1,\sigma_1 a_1,\sigma_2 \sigma_1 a_1,\ldots,\sigma_n \ldots \sigma_1 a_1)$ for some $\sigma \in Gal(K/k)^n$ to obtain the claim.
A: Reuns's answer is short and great. I'll give a down to earth proof so we can see what happenes at the level of elements.
As $K/k$ is finite Galois, it is finite and separable, by primitive element theorem, there exists $\alpha\in K$ with its minimal polynomial $f(x)$ over $k$ s.t $\frac{k[x]}{(f(x))}\cong K,x\mapsto \alpha$. And for each $\sigma\in G$, there exists a polynomial $f_\sigma (x)$ over $k$ s.t. $\frac{k[x]}{(f(x))}\to \frac{k[x]}{(f(x))},x\mapsto f_\sigma (x)$ correpsonds to $\sigma:K\to K$.
We first deal with the case $n=2$:
$$ \begin{align}
K\otimes_k K & \cong \frac{k[x_0]}{(f(x_0))}\otimes  \frac{k[x_1]}{(f(x_1))} \\ 
& \cong \frac{k[x_0,x_1]}{(f(x_0),f(x_1))}\\
& \cong \frac{K[x_0,x_1]}{(f(x_0),\prod_{\sigma\in G}(x_1-f_\sigma (x_0)))}\\
& \cong \prod_{\sigma\in G} \frac{\frac{k[x_0]}{(f(x_0))}[x_1]}{(x_1-f_\sigma (x_0))}\\
& \cong \prod_{\sigma\in G} \frac{k[x_0,x_1]}{(f(x_0),x_1-f_\sigma(x_0))}\\
& \cong \prod_{\sigma\in G} K\\
\end{align} $$
$$x_0\otimes x_1 \mapsto (x_0 f_\sigma (x_0))_{\sigma \in G}$$
$$\alpha(x_0)\otimes \beta(x_1)\mapsto (\alpha(x_0)\beta(f_\sigma(x_0)))_{\sigma\in G}$$
$$\alpha\otimes \beta\mapsto (\alpha \cdot\sigma(\beta))_{\sigma\in G}$$
If we tensor with another $K\cong \frac{k[x_2]}{(f(x_2))}$, let $s$ be any of the primitive element in $\frac{k[x_0,x_1]}{(f(x_0),x_1-f_\sigma (x_0))}$, it could be $x_0,f_\sigma(x_0)$ or any $f_\tau (x_0)$ where $\tau\in G$. Then we have
$$\frac{k[x_0,x_1]}{(f(x_0),x_1-f_\sigma (x_0))}\otimes_k \frac{k[x_2]}{(f(x_2))}\cong \prod_{\tau\in G}\frac{k[x_0,x_1,x_2]}{(f(x_0),x_1-f_\sigma (x_0),x_2-f_\tau(s))}\cong \prod_{\tau\in G}K$$
Then $x_0\otimes x_1 \otimes x_2$ sends to $x_0 \cdot f_\sigma (x_0) \cdot f_\tau (s)$, if we take $s=x_1=f_\sigma (x_0)$, we have $\alpha_0\otimes \alpha_1 \otimes \alpha_2 \mapsto \alpha_0 \cdot \sigma (\alpha_1)\cdot \tau \sigma (\alpha_2)$
In the general case, in
$${\bigotimes_k}^n K\cong \frac{k[x_0,\cdots,x_n]}{(f(x_0),...,f(x_n))}$$
given an element $(\sigma_1,...,\sigma_n)\in G^n$, the correspoindg $K$-factor is
$$\frac{k[x_0,\cdots,x_n]}{(f(x_0),x_1-f_{\sigma_1}(x_0),x_2-f_{\sigma_2}(x_1),\cdots,x_n-f_{\sigma_n}(x_{n-1}))}$$
$$\cong \frac{k[x_0,\cdots,x_n]}{(f(x_0),x_1-f_{\sigma_1}(x_0),x_2-f_{\sigma_2}f_{\sigma_2}(x_0),\cdots,x_n-f_{\sigma_n}\cdots f_{\sigma_1}(x_0))}\cong K$$
So $x_0\otimes\cdots \otimes x_n\mapsto x_0\cdot [f_{\sigma_1}(x_0)]\cdots [ f_{\sigma_n}\cdots f_{\sigma_1}(x_0)]$ which further induces the map of abstract elements of $K$,
$$\alpha_0\otimes\cdots \otimes \alpha_n\mapsto  \alpha_0 [\sigma_1(\alpha_1)][\sigma_2 \sigma_1(\alpha_2)]\cdots [(\sigma_n\cdots\sigma_1)(\alpha_n)]$$
The result follows. As we can see, the above form is not necessary, if we take the primitive element to be $x_0$, then we can get the following nice written identification:
$$\alpha_0\otimes\cdots \otimes \alpha_n\mapsto  \alpha_0 [\sigma_1(\alpha_1)][\sigma_2 (\alpha_2)]\cdots [\sigma_n(\alpha_n)]$$
