 Solution

Set $K_r=K({\alpha }_1,\dots ,{\alpha }_r)$, beginning with $K_0=K$. Suppose $j:K_{r-1}\to L$ extends $i$. The minimal polynomial of ${\alpha }_r$ over $K_{r-1}$ divides its minimal polynomial over $K$. Its image under $j$ therefore divides $i(m_{{\alpha }_r})$, which splits into distinct linear factors in $L$. It consequently has exactly $[K_r:K_{r-1}]$ distinct roots in $L$. Part (i) supplies exactly this many extensions of $j$ to $K_r$.

Starting with the unique prescribed map on $K$, multiply these branching counts:For the final assertion, let $M$ be a splitting field of a separable polynomial and embed it in an algebraic closure $\Omega$. It is generated by roots whose minimal polynomials are separable and split in $\Omega$. Applying the just-proved count with $L=\Omega$ gives $[M:K]$ embeddings. Were $M/K$ inseparable, part (a) would give strictly fewer. Thus the splitting extension is separable. This works for the factorwise convention as well, since the generators' minimal polynomials divide the separable irreducible factors.