How to Show $\prod_{n=1}^\infty\bigg(\frac{z^n}{n!}+e^{\frac{z}{2^n}}\bigg)$ Converges Uniformly on Compact Sets Prove that $$\prod_{n=1}^\infty\bigg(\frac{z^n}{n!}+e^{\frac{z}{2^n}}\bigg)\tag{$*$}$$ converges uniformly on compact sets to an entire function. 
I haven't seen a problem like this before, so I don't know what tricks/techniques are usually applied. If you know of any please share.
Attempt 1: (Partial Products) Given $M\in \mathbb{N}$, consider the compact disc $\Omega_M=\lbrace z\ \in \mathbb{C}:\vert z \vert \le M\rbrace$.
Define $P_N$ to be the $N$th partial product, so $$P_N =\bigg(z+e^{\frac{z}{2}}\bigg)\bigg(\frac{z^2}{n!}+e^{\frac{z}{2^2}}\bigg)\cdots\bigg(\frac{z^N}{N!}+e^{\frac{z}{2^N}}\bigg) \\=\frac{z^{1+2+\cdots + N}}{1!2!\cdots N!}+ze^{\bigg(\frac{z}{2^2}+\cdots+\frac{z}{2^N}\bigg)}+z^2e^{\bigg(\frac{z}{2}+\frac{z}{2^3}+\cdots+\frac{z}{2^N}\bigg)}\\+z^3\Bigg[e^{\bigg(\frac{z}{2}+\frac{z}{2^2}+\frac{z}{2^4}\cdots+\frac{z}{2^N}\bigg)}+e^{\bigg(\frac{z}{2}+\frac{z}{2^3}+\cdots+\frac{z}{2^N}\bigg)}\Bigg]+\cdots +e^{\bigg(\frac{z}{2}+\frac{z}{2^2}+\cdots+\frac{z}{2^N}\bigg)}$$
If I take absolute value of both sides and apply the triangle inequality to the right-most side, use the fact that $\vert z \vert \le M$, each term seems to converge, but I have infinitely many terms in the whole sum and I'm not sure if the entire sum converges. Even if an upper bound converged, that wouldn't necessarily imply that $(*)$ converges. So I think this may be a dead end.
Attempt 2: (Weierstrass Factorization Theorem) If we can show the zero set of $(*)$ is $\lbrace z_j\rbrace_{j\ge 1}$, where none of the $z_j$ are zero and this set has no limits points, then it is holomorphic. The problem here is that I don't know how to find the solution of $\frac{z^n}{n!}+e^{\frac{z}{2^n}}=0$ for each $n$ (or for any $n$ for that matter.) Also, I am not sure where compactness would come into play in this approach.
 A: Note $f_n(z) = \frac{z^n}{n!}e^{\frac{z}{2^n}}$. 
Let's check that $\sum (1-f_n)$ is convergent in $\mathcal{H}(\mathbb{C})$. The fact is that $(1 - f_n) = (1 - \frac{z^n}{n!} - e^{\frac{z}{2^n}})$. 
Just note that 
$$|1 - e^{\frac{z}{2^n}}| \leq \sum_{k=1}^{\infty}|\frac{z^k}{2^{nk}k!}| \leq \frac{1}{2^n}\sum_{k=1}^{\infty}\frac{|z|^k}{k!}$$
Take a compact $K$, let's say it's included in $D(0,M)$. As the function $g : z \to \sum_{k=1}^{\infty}\frac{z^k}{k!}$ is holomorphic (and does not depend on $n$), it is bouded by a constant $C$ on the compact $K$. Therefore, 
$$|1 - e^{\frac{z}{2^n}}| \leq C\frac{1}{2^n}$$
As a consequence of this, we have
$$\sum |1 - f_n(z)| \leq \sum_{n=0}^{\infty}\frac{M^n}{n!} + C \sum \frac{1}{2^n} \leq e^M + 2C$$
So the series is uniformly convergent on the compact $K$. Therefore, the product $\Pi f_n$ is also uniformly convergent on every compact set.
Here is the theorem I've used.

Theorem. Let $(f_n)$ a sequence of holomorphic functions. Suppose the serie $\sum (1 -f_n)$ is uniformly convergent in compact sets.
  Then, the sequence of holomorphic functions $(g_n)$ defined by 
  $$g_n =\prod_{k=0}^nf_k $$
  is uniformly convergent on compact sets. In particular, its limit $g =
\lim g_n$ exists, is noted $\prod_{n=0}^{\infty}f_n$, and defines a
  holomorphic function called the product of the $f_n$. Moreover, if
  neither $f_n$ takes the value $0$, so do $g$.

