Let h(x) be an irreducible factor of f(g(x)) in F[x]. Let u be a root of h(x). Then [F(u): F] = deg h(x) since h(x) is irreducible. Since h(x) is a factor of f(g(x)) and since u is a root of h(x), it follows that g(u) is a root of f(x). So [F(g(u)): F] = n since f(x) is an irreducible polynomial of degree n. Since F(g(u)) is contained in F(u), it follows that

[F(u): F] = [F(u): F(g(u))][F(g(u)): F].

So deg h(x) = [F(u): F] = [F(u): F(g(u))]n.