Let $\{a_1, \ldots, a_m\}$ be a basis of $K$ over $F$, and $\{b_1, \ldots, b_n\}$ a basis of $L$ over $K$. We show $\{a_i b_j : 1 \leq i \leq m,\, 1 \leq j \leq n\}$ is a basis of $L$ over $F$.

Spanning: Let $\ell \in L$. Since $\{b_j\}$ spans $L/K$: $\ell = \sum_{j=1}^{n} k_j b_j$ with $k_j \in K$. Since $\{a_i\}$ spans $K/F$: $k_j = \sum_{i=1}^{m} f_{ij} a_i$ with $f_{ij} \in F$. Therefore:

$\ell = \sum_{j=1}^{n} \left(\sum_{i=1}^{m} f_{ij} a_i\right) b_j = \sum_{i,j} f_{ij} (a_i b_j).$

Linear independence: Suppose $\sum_{i,j} f_{ij} (a_i b_j) = 0$ with $f_{ij} \in F$. Rewrite:

$\sum_{j=1}^{n} \left(\sum_{i=1}^{m} f_{ij} a_i\right) b_j = 0.$

Since $\{b_j\}$ is linearly independent over $K$, each inner sum vanishes: $\sum_{i=1}^{m} f_{ij} a_i = 0$ for each $j$. Since $\{a_i\}$ is linearly independent over $F$: $f_{ij} = 0$ for all $i, j$.

Therefore $\{a_i b_j\}$ is a basis with $mn$ elements, giving $[L:F] = mn$. $\blacksquare$