To prove that the norm of a tower of field extensions is the composition of norms We know that if we set $K$, $F$ and $L$ fields, with $L$ a finite extension of $F$ and $F$ a finite extension of $K$. Then we have the norm equality $$N_{L/K}=N_{F/K}\circ N_{L/F}$$ The common proof is by using rational canonical forms and elementary matrices. I was wondering if there is a way only by using the tools of field extensions. Thank you.
 A: The ideas presented in @SoumikGhosh 's answer apply rather to the case when (at least some of) the extensions are separable, so here is a more general argument. I will assume that you know some basic results on non-separable field extensions. 
First, let us fix an algebraic closure $\bar{K}$ of a field $K$ so that all finite extensions of $K$ are naturally seen as subfields of $\bar{K}$.
For a finite extension $F/K$, we denote by $[F:K]_i$ and $[F:K]_s$ respectively its inseparable and separable degrees. The set of $K$-fixing endomorphisms $\text{Hom}_K(F,\bar{K})$ has cardinality $[F:K]_s$. The separable and inseparable degrees are multiplicative in towers of finite field extensions. 
For $\alpha\in F$, the norm $N_{F/K}(\alpha)$ is defined as the determinant of the $K$-linear map $m_\alpha:F\to F$ given by multiplication by $\alpha$. If $P$ is the minimal polynomial of $\alpha$ over $K$, having $r_1,...,r_d$ as roots (counting multiplicities), then $$N_{F/K}(\alpha)=\left(\prod_{k=1}^dr_k\right)^{[F:K(\alpha)]}.$$ To prove the latter you will need to do some linear algebra: observe that $m_\alpha$ stabilizes the subspace $K(\alpha)$ and the restriction to the subspace can be represented by the companion matrix of $P$; and use that the determinant is the product of the eigenvalues, which, in this case, are the roots of $P$. 
Let $\text{Hom}_K(F,\bar{K})=\{\sigma_1,...,\sigma_s\}$. Using the expression from the last paragraph, one can prove that for any $\alpha\in F$, $$N_{F/K}(\alpha)=\prod_{j=1}^{[F:K]_s}\sigma_j(\alpha)^{[F:K]_i}$$ by observing that each root of the minimal polynomial of $\alpha$ over $K$ has multiplicity $[K(\alpha):K]_i$ and each $K$-embedding $K(\alpha)\to \bar{K}$ can be extended to a homomorphism $F\to\bar{K}$ in exactly $[F:K(\alpha)]_s$ different ways. 
Now suppose that $L/F$ is another finite extension and let $\text{Hom}_F(L,\bar{K})=\{\tau_1,...,\tau_t\}$. Fix any $\alpha\in L$. Extend each $\sigma_j$ to a $K$-embedding $\sigma_j':F(\tau_1(\alpha),...,\tau_t(\alpha))\to \bar{K}$. Then, using the above results, $$N_{F/K}\circ N_{L/F}(\alpha)=\prod_{j=1}^s\prod_{l=1}^t\sigma_j'(\tau_l(\alpha))^{[L:K]_i}.$$ Note that since $N_{L/F}(\alpha)\in F$, the choices of $\sigma_j'$ do not have any influence. 
The embeddings $\sigma_j'\circ\tau_l:L\to\bar{K}$ fix the subfield $K$ and are all different. Indeed, if $\sigma_{j_1}'\circ\tau_{l_1}=\sigma_{j_2}'\circ\tau_{l_2}$, then restricting to $F$ gives $j_1=j_2$ and then necessarily $l_1=l_2$. Since $[L:K]_s=st$, these compositions give all the elements of $\text{Hom}_K(L,\bar{K})$. We finish by again using the norm expression for $N_{L/K}(\alpha)$. 
A: let $K,F,L$ be field extensions in that order and $M$ a Galois extension over $K$ containing $M$. All extensions are assumed to be finite. Let $U_1$ be a set of coset representatives of $Gal(M/L)$ in $Gal(M/F)$ and $U_2$ be a set of coset representatives of $Gal(M/F)$ in $Gal(M/K)$. Then we have $N_{F/K}N_{L/F}(\alpha)=\prod_{\tau \in U_2} (\prod_{\sigma \in U_1} \tau\sigma(\alpha))=\prod_{\tau_j,\sigma_i} \tau_j\sigma_i(\alpha)$. But $\tau_j\sigma_i$ form a coset representative of $Gal(M/L)$ in $Gal(M/K)$ and hence the R.H.S is equal to $N_{L/K}(\alpha)$. Hope this avoids too much mess with matrices.
A: We prove the theorem for separable field extensions $F$ over $K$ and $L$ over $F$. By the theorem of the primitive element in field theory there exists a monic irreducible polynomial $f(x)=x^m+a_{m-1}x^{m-1}+\dots+a_0$ such that
$$F=K[x]:=K[X]/\langle f(X)\rangle.$$
Similarly there exists a monic irreducible polynomial $g(y)=y^n+b_{n-1}y^{n-1}+\dots+b_0$ with coefficients $b_k\in F$ such that
$$L=K[x,y]=F[y]:=F[Y]/\langle g(Y)\rangle.$$
The field $L$ has dimension $m\cdot n$ over the field $K$. By the theorem of the primitive element there exists a monic irreducible polynomial $h(z)=z^{mn}+c_{mn-1}z^{m-1}+\dots+c_0$ with coefficients $c_k\in K$ such that
$$L=K[z]:=K[Z]/\langle h(Z)\rangle.$$
Every zero $z_i$ of the polynomial $h(z)$ produces an isomorphic field
$$L\cong L_i=K[z_i]:=K[Z_i]/\langle h(Z_i)\rangle$$
and therefore there exists an isomorphism $\hat\sigma_i$ such that $L_i=\hat\sigma_i(L)$ with $z_i=\sigma_i(z)$. But because the field $L$ is also generated by adjoining the zeros $x\in K$ and $y\in F$
$$L=K[x,y]$$
this homomorphism $\hat\sigma_i$ must send the zero $x$ to a zero $x_i:=\hat\sigma_i(x)$ of the polynomial $f(x)$ and the zero $y$ to a zero $y_{ik}:=\hat\sigma_i(y)$ of the polynomial 
$$g_i(y):=\hat\sigma_i(g(y))=y^n+\hat\sigma_i(b_{n-1})y^{n-1}+\dots+\hat\sigma_i(b_0).$$
For every $i$ there are $n$ zeros $y_{ik}$ because the polynomial $g_i(y)$ remains irreducible over the field $L_i$. We define the homomorphism $\sigma_{ik}$ over the field $L$ to be the homomorphism that sends the zero $x$ to the zero $x_i$ and the zero $y$ to the zero $y_{ik}$ that was adjoined over the field $K[x_i]$:
$$L_{ik}:=K[x_i,y_{ik}]=\sigma_{ik}(K[x,y]).$$
In adjoining the zero $y_{ik}$ to the fields $K[x_i]$ there exists $n$ homomorphism $\tau_{ik}$ such that
$$L_{ik}:=K[x_i,y_{ik}]=\tau_{ik}(K[x_i,y_{i0}])$$
and the field $F=K[x_i]$ remains invariant. Now we have developed the means to prove the formula in the question. The norm of an element $\zeta\in L$ is defined to
$$\tag{1}N_{L/K}(\zeta):=\prod_{0\le i\lt mn}\hat\sigma_{i}(\zeta)=\prod_{0\le i\lt m}\prod_{0\le k\lt n}\sigma_{ik}(\zeta).$$
We reorder the zeros such that $x_0=x$. Let the homomorphism $\sigma_{0,k}$ be the identity on the zero $x$. Each factor in the product $\prod_{0\le k\lt n}\sigma_{0k}(\zeta)$ contains exactly one of the zeros $y_{0,k}$ of the polynomial $g(y)$ that can be adjoined above to the field $F$. But then this product must be a symmetrical polynomial in the zeros $y_{0,k}$ and we can write the product to
$$\tag{2}\prod_{0\le k\lt n}\sigma_{0k}(\zeta)=d_mx^m+d_{m-1}x^{n-1}+\dots+d_0$$
with coefficients $d_j\in K[x,S_{00},\dots,S_{0m}]$ with the symmetrical polynomials
$$S_{00}:=1, S_{01}:=y_{01}+\dots+y_{0n}, \dots, S_{0n}:=y_{01}\cdots y_{0n}$$
and $y_{0k}:=\tau_{0k}(y)$. But these symmetric polynomials are known to be $S_{0k}=(-1)^kb_k\in F=K[x]$ of the polynomial $g(y)$ above. Therefore the product $(2)$ gives an element in the field $F=K[x]$.
In taking the product $\prod_{0\le k\lt n}\sigma_{ik}(\zeta)$ for a fixed $i$ we get a similar  result to equation $(2)$ with 
$$\tag{3}\prod_{0\le k\lt n}\sigma_{ik}(\zeta)=\sigma_{ik}(d_m)x_i^m+\sigma_{ik}(d_{m-1})x_i^{n-1}+\dots+\sigma_{ik}(d_0)$$
with coefficients $\sigma_{ik}(d_j)\in K[x_i,\sigma_{ik}(S_0),\dots,\sigma_{ik}(S_m)]$. But those are just the coefficients of the poynomial $g_i$ or $S_{ik}=\sigma_{ik}(S_{0k})=(-1)^k\sigma_{ik}(b_k)\in K[x_i]$.
The homomorphism $\sigma_{ik}$ of the field $L$ over $K$ can be restricted to the homomorphism $\kappa_i$ of the field $K[x_i]$ over the field $K$. But these homomorphism $\kappa_i$ just map the the zero $x$ to the zero $x_i$ and therefore they are the field homomorphism of the field $F=K[x_i]$ over the field $K$. In the formulas $(1)$ and $(3)$ we have seen that the inner product over $k$ gives an element in the field $K[x_i]$. Because the formulas $(2)$ and $(3)$ give
$$\prod_{0\le k\lt n}\sigma_{ik}(\zeta)=\kappa_i\left(\prod_{0\le k\lt m}\sigma_{0k}(\zeta)\right)$$
we can rewrite the formula $(1)$  to
$$N_{L/K}(\zeta)=\prod_{0\le i\lt m}\kappa_i\left(\prod_{0\le k\lt n}\sigma_{0k}(\zeta)\right).$$
But the homomorphism $\sigma_{0k}$ are the homomorphism $\tau_k:=\tau_{0,k}=\sigma_{0k}$ of the field $L$ over the field $F$. This gives directly
$$N_{L/K}(\zeta)=N_{F/K}\circ N_{L/F}(\zeta)=\prod_{0\le i\lt m}\kappa_i\left(\prod_{0\le k\lt n}\tau_{k}(\zeta)\right).$$
