Let \(k\subseteq F\subseteq L\) be field extensions with \([L:F]\) and \([F:k]\) finite. Then \([L:k]\) is finite and \[ [L : k] = [L: F]\,[F: k]. \]
Let \(\alpha_1,\ldots,\alpha_m\) be a basis of \(F\) over \(k\) and \(\beta_1,\ldots,\beta_n\) a basis of \(L\) over \(F\). Every \(x\in L\) is \(\sum_j c_j\beta_j\) with \(c_j\in F\), and each \(c_j=\sum_i a_{ij}\alpha_i\) with \(a_{ij}\in k\), so the \(\alpha_i\beta_j\) span \(L\) over \(k\). If \(\sum_{i,j}a_{ij}\alpha_i\beta_j=0\) with \(a_{ij}\in k\), then linear independence of the \(\beta_j\) over \(F\) gives \(\sum_ia_{ij}\alpha_i=0\) for each \(j\), and linear independence of the \(\alpha_i\) over \(k\) gives \(a_{ij}=0\). So the \(mn\) products \(\alpha_i\beta_j\) form a basis of \(L\) over \(k\).