= Solution
The <tower law for field extensions> states that for finite <field extension>[field extensions]
$$
K\subseteq E\subseteq L,
$$
one has
$$
\boxed{[L:K]=[L:E][E:K]}.
$$
To prove it, let $(u_i)_{i=1}^m$ be a $K$-basis of $E$ and $(v_j)_{j=1}^n$ an $E$-basis of $L$. Every $x\in L$ can be written as $\sum_j a_jv_j$ with $a_j\in E$, and expanding each $a_j$ in the basis $(u_i)$ shows that the $mn$ products $u_iv_j$ span $L$ over $K$.
If
$$
\sum_{i,j}c_{ij}u_iv_j=0,\qquad c_{ij}\in K,
$$
then independence of the $v_j$ over $E$ gives $\sum_i c_{ij}u_i=0$ for every $j$, and independence of the $u_i$ over $K$ gives every $c_{ij}=0$. Thus $(u_iv_j)$ is a $K$-basis of $L$, proving the formula.
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