# dimension of vector space

Let $E,F,K$ be fields such that $F \subset E \subset K$. Prove that if $A=\{a_1,...,a_n\}$ is a basis for $K$ over $E$ and $B=\{b_1,...b_m\}$ is a basis for $E$ over $F$, then $C=\{a_ib_j, 1 \leq i \leq n, 1 \leq j \leq m\}$ is a basis for $K$ over $F$. Conclude that if $\dim_E(K)$ and $\dim_F(E)$ are finite, then $\dim_F(K)=\dim_E(K)\dim_F(E)$.
I know how to do the proof, but I'm not sure how to conclude that $\dim_F(K)=\dim_E(K)\dim_F(E)$. Some help please.
$\dim_E(K)=n,\dim_F(E)=m,\dim_F(K)=n*m,\longrightar row \dim_F(K)=\dim_E(K)\dim_F(E)\ ?