You can use that is analytic on if and only if there exists an open set containing such that

there exists holomorphic on extending . Then the assertion follows from the fact that sums and products of holomorphic functions are holomorphic.

Also you can use that is analytic on if and only if for each compact subset

of there exists such that for each . Here the sum is almost immediate but for the product you will need Leibnitz rule.

If you want a direct proof for the product, puff, let say using Cauchy Series could be done, but anyway you will need the fact that every analytic expansion converges absolutely, which follows from the characterization above.