= Solution
Set $K_r=K(\alpha_1,\ldots,\alpha_r)$, 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:
$$
\boxed{\#\{\widetilde\jmath:\widetilde K\to L,\ \widetilde\jmath|_K=i\}
=\prod_r[K_r:K_{r-1}]=d.}
$$
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 \b[the splitting extension is separable]. This works for the factorwise convention as well, since the generators' minimal polynomials divide the separable irreducible factors.
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