Zeros in the complex plane and convergence I'm doing some number theory which requires some work in $\mathbb{C}$, but unfortunately my complex analysis is a little rusty. 
A text I am reading states the following:

...and given that $\sum_{\rho}|\rho|^{-2}$ converges, it follows that the product $\prod \limits_\rho (1-\frac{z}{\rho})e^{z/\rho}$ converges to an entire function with zeros at $z=\rho$ and nowhere else.

Could anyone explain, preferably without going into too much detail (just refreshing my memory!) why this is? Here, $\rho$ ranges over the set of zeros of an integral function $f: \mathbb{C} \to \mathbb{C}$ of order 1, which is not identically zero. It's obvious that the product is zero at each $\rho$; what do we need to know it converges to an entire function with zeros nowhere else?
For convergence, do we simply need to bound the modulus above by some finite value at each $z$? And since it's an infinite product, how do we confirm that it's zero nowhere else; a lower bound of the same manner? Finally, what tells us it is entire? Again, given that it's an infinite product I don't know how trivial that claim is; not very hard I'm sure but perhaps requiring some small justification. 
As I said, I have probably learned all this at some point in my life so no need for extraordinary detail, just a brief explanation would be very helpful. It all seems vaguely reminiscent of a course I took on Riemann surfaces a few years back. Thanks!
 A: Just for ease of notation, enumerate the zeros as $(\rho_n)$. Then for any $R>0$ there exists $n_R$ such that $|\rho_n| > 2R$ for $n\ge n_R$. For any $N > n_R$, the function 
$$g_N(z) = \prod_{n=n_R}^N \left(1-\frac{z}{\rho_n}\right) e^{z/\rho_n}$$
has no roots in $|z|<R$, so there exists an analytic branch of
$$ h_N(z) = \log g_N(z) = \sum_{n=n_R}^N \left[ \log\left(1-\frac{z}{\rho_n}\right) + \frac{z}{\rho_n}\right]
$$
where we choose the principal branch of the logarithm for each term in the sum. Using the fact that $\left|\frac{z}{\rho_n}\right|<\frac12$ for $|z|<R$ and $n\ge n_R$, and the fact that there exists a constant $C$ such that for $|w| <\frac12$ we have $|\log(1-w)+w| \le C|w|^2$, again for the principal branch of the logarithm (which follows easily from the power series expansion), we get that
$$
\left| \log\left(1-\frac{z}{\rho_n}\right) + \frac{z}{\rho_n}\right| \le C \frac{|z|^2}{|\rho_n|^2} \le \frac{CR^2}{|\rho_n|^2}
$$
for all $n \ge n_R$ and $|z|<R$. By assumption the series over those terms converges, so 
$$
\lim_{N\to\infty} h_N(z) = \sum_{n=n_R}^\infty \left[ \log\left(1-\frac{z}{\rho_n}\right) + \frac{z}{\rho_n}\right]
$$
converges uniformly on $|z|<R$ to an analytic limit $h(z)$. This implies that
$$
\lim_{N\to\infty} g_N(z) = \lim_{N\to\infty} e^{h_N(z)} =  e^{h(z)} = \prod_{n=n_R}^\infty \left(1-\frac{z}{\rho_n}\right) e^{z/\rho_n}
$$
is a non-zero analytic function on $|z|<R$. This shows that
$$
f(z) = \prod_{n=1}^\infty \left(1-\frac{z}{\rho_n}\right) e^{z/\rho_n}
$$
converges uniformly to an analytic function on $|z|<R$, whose only zeros are $z=\rho_n$ with $n<n_R$ and $|\rho_n|<R$. Since $R$ was arbitrary to begin with, the claim follows, and the convergence of the infinite product is locally uniform in the whole plane.
