Help me understand this missing detail: transitivity of the field norm. A very basic fact about norms on field extensions is the following:

Suppose $E/F/k$ is a tower of finite extensions. Then, 
  $ N_{E/k}(\alpha) = N_{F/k}(N_{E/k}(\alpha)), $ for all $\alpha \in E$.

There are various proofs of this fact, but I think I should be able to make the following simple proof work. I have almost all the details, just one thing is missing: 
Proof: Let $m = [E:F]$ and $n = [F:k]$ and fix the ordered bases $\{e_1, \dots, e_m\}$ and $\{f_1, \dots, f_n\}$ of $E/F$ and $F/k$, respectively. Thus a basis for $E/k$ is $\{e_i f_j\}$. Suppose that $\alpha \in F$. Then consider its multiplication map, 
$F \overset{\cdot \alpha}{\to} F$, which has the effect $\alpha f_j = \sum_{i=1}^n c_{ij} f_i$. Thus $(c_{ij})$ denotes the matrix of $\alpha$ as a multiplication map on $F/k$ with respect to the ordered basis given. Now observe that the effect on the basis of $E/k$ is $\alpha e_i f_j = \sum_{k} c_{kj} f_k e_i$, so the matrix of $\alpha$ over $E/k$ is block diagonal with $m$ blocks, $(c_{ij})$. It is immediate that 
$$ N_{E/k} = \det((c_{ij}))^m = N_{F/k}(\alpha)^{[E:F]}.$$
Now suppose $\alpha \in E$. From of our previous discussion it follows at once that $N_{E/k}(\alpha) = N_{F(\alpha)/k}(\alpha)^{[E:F(\alpha)]}$. It suffices now to show that 
$$
N_{F/k}(N_{F(\alpha)/F}(\alpha)) \overset{!}{=} N_{F(\alpha)/k}(\alpha).
$$
If it were true that ! holds, then 
$$ 
N_{E/k}(\alpha) = N_{F(\alpha)/k}(\alpha)^{[E:F(\alpha)]} = 
N_{F/k}(N_{F(\alpha)/F}(\alpha))^{[E: F(\alpha)]} = 
N_{F/k}(N_{E/F}(\alpha))$$
The penultimate equality comes from ! and the final inequality from our previous discussion. Note that the minimal polynomial for $\alpha$ over $F$ is $f(T) \in F[T]$, which is $f(T) = T^d + \sum_0^{d-1} a_k T^k$, for 
$a_k \in F$. Thus, $N_{F(\alpha)/F}(\alpha) = (-1)^d a_0$. Thus, 
$$
N_{F/k}(N_{F(\alpha)/F}(\alpha)) = N_{F/k}((-1)^d a_0) = 
(-1)^{dn} \det(A_0),$$
where $A_0$ is the matrix for the multiplication of $a_0$. Thus it simply remains to see that 
$$
(-1)^{dn} \det(A_0) =  N_{F(\alpha)/k}(\alpha).
$$
Question: Why does the equality above hold?
 A: Observe that a basis for $F(\alpha)$ is $\{f_i \alpha^{j-1}\}$, where $1 \leq i \leq n$ and $1 \leq j \leq d$. For each of the $a_i$ appearing in $f(T)$ above, associate $A_i$, the matrix of the multiplication map of $a_i: F \to F$. Now compute the matrix for $\alpha: F(\alpha) \to F(\alpha)$. Its effect on a basis element is $\alpha f_i \alpha^{j-1} = f_i \alpha^{j}$. If $j = n$, then $f_i \alpha^j = f_i (-a_0 -a_1\alpha - \cdots -a_{n-1} \alpha^{n-1})$. Thus the matrix for $\alpha$ as a multiplication map on $F(\alpha)$, is (with respect to the given basis):
\begin{bmatrix}
0 & 0 & 0 & \cdots & -A_0\\  
I_n & 0 & 0 & \cdots & -A_1\\
0 & I_n & 0 & \cdots & -A_2 \\
\vdots & & \ddots & & \vdots\\ 
0 & 0 & \cdots & I_n & -A_{d-1} \\
\end{bmatrix}
We can cycle the last column to the right, which causes $dn - 1$ transpositions, and needs to be done $n$ times, giving that the determinant of the matrix above is $$(-1)^{n(nd-1)}\det(-A_0) = (-1)^{n^2d}\det(A_0) = 
\det(A_0)(-1)^{nd},$$ as desired.
