Degree of Field Extensions is Multiplicative

Theorem
Let $$E/K$$ and $$K/F$$ be finite field extensions. Then $$E/F$$ is a finite field extension and $$[E:F]=[E:K][K:F]$$

Proof
First, note that $$E/F$$ is a field extension as $$F \subseteq K \subseteq E$$.

Suppose $$[E:K]=m$$, $$[K:F]=n$$. Let $$\alpha = \{a_1,...,a_m\}$$ be a basis of $$E/K$$ and $$\beta = \{b_1,...,b_n\}$$ be a basis of $$K/F$$. We wish to prove the set

$$\gamma = \{a_ib_j~:~1\leq i \leq m,~1\leq j \leq n\}$$

is a basis of $$E/F$$. As $$\alpha$$ is a basis of $$E/K$$, then $$\forall ~ c\in E$$, we have

$$c= \sum_{i=1}^{m}c_ia_i$$, for some $$c_i \in K$$.

Letting $$b=\sum_{j=1}^{n}b_i$$ and $$\frac{c_i}{b}=d_i$$ (note $$c_i,b \in K \rightarrow d_i \in K$$ and $$b \neq 0$$ as $$\beta$$ is linearly independent over $$F$$), we have:

$$c = \sum_{i=1}^m \frac{c_i}{b}\cdot b\cdot a_i~=~\sum_{i=1}^{m}\sum_{j=1}^{n}d_ia_ib_j$$.

Thus $$\gamma$$ is a spanning set of $$E/K$$.

To show $$\gamma$$ is linearly independent, we first observe that as $$\beta \subseteq E$$, we can express each $$b_j$$ as $$b_j = \sum_{i=1}^m d_{ij}a_i$$ for some $$d_{ij} \in K$$. Hence we have

$$0 = \sum_{i=1}^m \sum_{j=1}^n c_{ij}a_ib_j$$

$$= \sum_{i=1}^m \sum_{j=1}^n \left(c_{ij}\sum_{i=1}^m d_{ij}a_i\right)a_i$$

Now, as $$\beta$$ is linearly independent, $$b_j \neq 0~\forall~j$$, hence $$\sum_{i=1}^m d_{ij}a_i \neq 0$$. Therefore, as all fields are integral domains, we have

$$0 = \sum_{i=1}^m \sum_{j=1}^n \left(c_{ij}\sum_{i=1}^m d_{ij}a_i\right)a_i \leftrightarrow \left(c_{ij}\sum_{i=1}^m d_{ij}a_i\right) = 0~\forall~i,j$$ by the linear independence of $$\alpha$$ over $$K$$

$$\leftrightarrow c_{ij}=0~\forall ~i,j$$ for the reasons mentioned above.

Hence $$\gamma$$ is a linearly independent spanning set; thus it is a basis.

Recalling the defintion of $$\gamma$$ as $$\{a_ib_j~:~1\leq i \leq m,~1\leq j \leq n\}$$, we have $$|\gamma |=mn=[E:K][K:F]$$, as desired.