 Solution

Put $L_i=K({\alpha }_1,\dots ,{\alpha }_i)$, with $L_0=K$, and letFix a $K$-embedding $\sigma :L_{i-1}\to \overline{K}$. If $m_i(t)$ is the minimal polynomial of ${\alpha }_i$ over $L_{i-1}$, an extension of $\sigma$ to $L_i$ is determined by the image of ${\alpha }_i$, which must be a root of the polynomial obtained by applying $\sigma$ to the coefficients of $m_i$. Conversely, each distinct root gives one extension. Since $\overline{K}$ is algebraically closed, there is at least one such root, and there are at most $d_i$ distinct roots.

Starting from the unique embedding of $K$, induction therefore givesIf $L/K$ is separable, each $m_i$ is separable, so every transformed polynomial has exactly $d_i$ distinct roots. Every inequality is then an equality. Taking $i=n$ proveswith equality in the separable case.