Understanding the roots in the Weierstrass factorization theorem To construct the Weirstrass product, we first start with something of the form $$\prod_{k=1}^\infty(z-a_k)$$
The next thing we do is include exponential factors out front (which obviously don't affect roots) and including this will ensure convergence.
Next we pull the roots at zero out to get something like...
$$z^m e^{f(z)} \prod_{k=1}^\infty(1-z/a_k)$$
It turns out that factoring the roots at $(1-z/a_k)$ is better than $(z-a_k)$ but I cannot see the reason why. They are both $0$ when $z = a_k$. Can someone explain to me, or prove to me, why factoring them in that way is preferable?
 A: First of all, the exponential factors included for convergence are not "out front" as you say.  Instead, the exponential factors are inside the product, so that they actually do matter for its convergence.  That is, the product has the form $$z^m e^{f(z)}\prod\left((1-z/a_k)\cdot(\text{exponential factor})\right).$$
Now, why can't we instead write it as $$z^m e^{f(z)}\prod\left((z-a_k)\cdot(\text{exponential factor})\right)?$$  The answer is that the types of exponential factors used in the statement of the theorem are chosen specifically so that the expressions $(1-z/a_k)\cdot(\text{exponential factor})$ will converge to $1$ as $k\to\infty$ (and by tweaking the parameters in the exponential factors, we can arrange that they converge to $1$ fast enough so that the product of infinitely many of them converges).  Of course, this is just a somewhat arbitrary convention, and you can just add a factor of $1/a_k$ to what you are calling the "exponential factors" and then $(z-a_k)\cdot(\text{exponential factor})$ will converge to $1$ instead.  So, you can get a factorization like $$z^m e^{f(z)}\prod\left((z-a_k)\cdot(\text{exponential factor})\right),$$ but the exponential factors will have slightly different form than in the usual statement of the theorem.
In particular, the usual statement of the theorem has the advantage that if $a_k\to \infty$ fast enough, then all the exponential factors can be taken to be just $1$, and so you get a nicer-looking factorization $$z^m e^{f(z)}\prod(1-z/a_k).$$  If you use factors of $z-a_k$ instead, then the exponential factors in that case will instead be $1/a_k$ so the factorization becomes $$z^m e^{f(z)}\prod\left((z-a_k)\cdot(1/a_k)\right).$$  Note that if there are infinitely many roots, then you definitely cannot hope to ever have a factorization like $$z^m e^{f(z)}\prod(z-a_k)$$ with no exponential factors since $z-a_k$ will go to $\infty$ and the product will diverge.
