Factorization of Taylor series. I know that for a (finite) polynomial $P(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_0$ whose zeros are $x_1, x_2, \ldots, x_n$, then we can factorize it as $$P(x) = a_n(x - x_1)(x - x_2) \cdots (x - x_n).$$
But does this apply to infinite series like the Taylor series?
For example, we know that $$\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \cdots$$ and the zeros of this function is $x = k \pi$, where $k \in \Bbb Z$, then does it hold that $$x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \cdots = x \prod_{k = 1}^{\infty} [(x - k \pi)(x + k \pi)]$$ and why?
 A: Yes you can do that for entire functions. https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem
There are a few things to keep in mind. They way you multiply all the roots, there are issues with convergence in it. To deal with it, they define functions $E_n(z)$ (see the definition in wikipedia), with the property that it is zero only for $z = 1$ and $\prod_{n} E_n(z)$ converges for $|z| <1$. 
After that its quite clear what to do. Let $z_n$ be the non-zero roots of the entire function $f$. If there are infinitely many of them, write them in a way such that $|z_n| \to \infty$ as $n \to \infty$. (This is possible as zeros of an entire function is discrete.)
Now consider $E_n (z/z_n)$. This has a zero only at $z= z_n$. So consider $\prod_n E_n (z/z_n)$. For any $z$, as $n$ becomes large we have $|z/z_n|  <1- \delta$ for some $\delta > 0 $ So by previous discussion, $\prod_{n} E_n (z/z_n)$ converges.
So far, we have not really considered the root zero. Let its multiplicity be $m \ge 0$. So, to include that root multiply the above factor by $z^m$ to get $z^m \prod_{n} E_n (z/z_n)$. Now consider $$\frac{f}{z^m \prod_{n} E_n (z/z_n)}.$$ All its singularities are removable. So it extends to an entire functions. But that entire function can not have any zeros. So  $$\frac{f}{z^m \prod_{n} E_n (z/z_n)} = e^g$$ for some entire function $g$. This gives you the required result!
