Elimination of zeros of an entire function

Question

Let be $f$ an entire function and $a_1,a_2,...,a_N$ the zeros of $f$ (i.e. $f(a_k)=0$). A $g$ function as: $$g(z)=\begin{cases} \frac{f(z)}{(z-a_1)(z-a_2)...(z-a_N)}, \qquad \text{for }z\neq a_k, \\ \lim_{z\rightarrow a_k}g(z), \qquad \text{for }z=a_k \end{cases}$$ where those limits exist. Show that $g$ is an entire function. I know that $\frac{f(z)-f(a)}{z-a}$ is an entire function.

Answer 1

Obviously the only points in which we may have problems are $a_1,...,a_N$. But it is easy to see that all these points are just removable singularities. For example, $a_1$ is zero of $f$ and by a known theorem in complex analysis we can write $f(z)=(z-a_1)h(z)$ when $h$ is holomorphic in the neighborhood of the point $a_1$. Can you finish from here?