Understanding the elementary factors in Weierstrass factorization theorem An infinite product such as $\,\prod _{n}(z-c_{n})$ cannot converge. In order for it to converge, each factor $(z-c_{n})$ must approach 1 as $n\to \infty$. So it stands to reason that one should seek a function that could be 0 at a prescribed point, yet remain near 1 when not at that point and furthermore introduce no more zeroes than those prescribed.
This is where Weierstrass' elementary factors come in as they have these properties and serve the same purpose as the factors $(z-c_{n})$ above.
So far I can follow, but then the elementary factors are presented as...
$E_{n}(z)={\begin{cases}(1-z)&{\text{if }}n=0,\\(1-z)\exp \left({\frac  {z^{1}}{1}}+{\frac  {z^{2}}{2}}+\cdots +{\frac  {z^{n}}{n}}\right)&{\text{otherwise}}.\end{cases}}$
and it is stated that the utility of the elementary factors $E_{n}(z)$ lies in the following lemma...

Lemma (15.8, Rudin) for $| z | ≤ 1$, $n\in \mathbb {N}$
$\vert 1-E_{n}(z)\vert \leq \vert z\vert ^{{n+1}}$

I simply do not understand why the Lemma is important, or even what it is trying to prove, nor do I understand the value of using the $exp$. There's no explanation surrounding it, so seeing it for the first time makes little sense to me.
Can someone please re-explain the Lemma in terms that might help me understand it, and perhaps prove to me why it is necessary to prove that $E_{n}(z)$ converges?
 A: I would recommend getting a good, purely Complex Analysis such as Ahlfor's for more details (at least do not start with Rudin) but the basic idea is as such:
To construct the Weirstrass product, we wish to get something of the form $$\prod_{k=1}^\infty(z-a_k)$$ Which is sometimes possible, but not always possible. We have to do some modifications. The first thing we do is include exponential factors out front (which obviously don't affect roots) and pull the roots at zero out to get something of the form
$$z^m e^{f(z)} \prod_{k=1}^\infty(1-z/a_k)$$
We now want to ensure the thing inside the product will actually converge. The classic way to ensure convergence without changing the location of zeros is to multiply by exponential factors, but we still want to preserve the product.
To make this work, we multiply by an exponential function whose exponent is chosen to be part of the Taylor Series for the natural logarithm. By a more general version of the argument
$$(1-z) = e^{\log(1-z)} = \exp \left( -\frac{z^1}{1}-\frac{z^2}{2}-\frac{z^3}{3}-\cdots \right)$$
we see that choosing our exponential factors to be of the form
$$\exp \left( \frac{z^1}{1}+\frac{z^2}{2}+\frac{z^3}{3}-\cdots + \frac{z^n}{k}\right)$$
Will result in a product with the same roots that is essentially of the form (for $|z| \le 1$)
$$z^m e^{f(z)}\prod_{k=1}^\infty (1 - \mathcal{O}(z^{k+1}/a_k))$$
With the appropriate definition of this product at $z = 0,$ we now have an idea of what we want our product to look like. We now have to see if this actually makes sense for $z$ outside the unit disk, and check if the estimates still work. The rest is a bunch of analysis to get the best results we can relating to this.

The Lemma you cite is one way of showing that via a slight modification of the above argument, we can in fact approach $1$ as fast as we need to via the right exponential factors, so that we can find such a product for any analytic function regardless of where the zeros are located.
Key terms to look up include:
Genus: the smallest integer $n$ such that $\sum\frac{1}{|a_k|^{n+1}}$ converges. This gives us a basic idea how fast the roots spread out
Order, Type: slightly more technical definitions telling us how fast our zeros spread out by comparing them to (hyper)exponential functions, which again tells us a lot about how nice our product will be
