Order of an entire function represented as an infinite product Question- What is the order of growth of the entire function given by the infinite product of $1-(z/n!)$ where $n$ goes from 1 to infinity?
My thoughts- I have already proven that the infinite product converges to an entire function. One can see that the zeros of the infinite product are at $n!$.
So, by Hadamard's factorization theorem, I can deduce that the order $p$ is such that $0\leq p <1$.
I suspect the order of growth is zero, but using inequalities like 
$\lvert \log(1+z)\vert < \lvert z\rvert$ for $\lvert z\rvert < 1$ and $e^n < n!$ for $n\geq 6$, all I could get was the order of growth is less than $1$.
I feel like there should be some simple inequality argument that shows us the order is zero.
Can somebody give me a hint?
Thanks.
 A: Claim: $p = 0$
Let $f(z) = \prod_{i=1}^\infty (1-\frac{z}{n!})$
I will show that for any $\epsilon >0$, the order of growth $p \leq \epsilon \;$ i.e. I will show $\exists \; A, B >0$ such that $|f(z)| \leq Ae^{B|z|^\epsilon}$.
$\begin{align}|f(z)| = |\prod_{i=1}^\infty (1-\frac{z}{i!})| &= |\prod_{i! \leq |z|} (1-\frac{z}{i!})| \;|\prod_{i! > |z|} (1-\frac{z}{i!})| \\
&\leq \; \prod_{i! \leq |z|} (1+\frac{|z|}{i!}) \; \prod_{i! > |z|} (1+\frac{|z|}{i!}) \end{align}$ 
Let us fix $z$, and suppose $\;n! \leq |z| < (n+1)!$ and $n\geq 6$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ (1)
Then, we use the inequality $e^n < n!$ for $n>6$ to get $e^n \leq |z| \implies n \leq log|z|$ $\;\;\;\;\;\;\;\;\;$ (2)
Now,
$\begin{align}\prod_{i! \leq |z|} (1+\frac{|z|}{i!}) &\leq \prod_{i! \leq |z|} (\frac{2|z|}{i!}) \;\;\;(since \;\;\frac{|z|}{i!}\geq 1) \\ & \leq\prod_{i! \leq |z|} |z| \\
& \leq |z|^n \;\;\;\;\;\;\;\;\;\;\;(since \;\; n! \leq |z|) \\
& \leq |z|^{log|z|} \;\;\;\;\;\;\;\; by \;\mathbf{(2)} \\
\end{align}$
Note that for any $\epsilon > 0, \frac{(log|z|)^2}{|z|^\epsilon} \rightarrow 0$ as $|z| \rightarrow \infty$.
Thus, $\exists \; C > 0, D > 0$ such that 
$$(log|z|)^2 \leq C + D|z|^{\epsilon} \implies |z|^{log|z|} \leq e^{C +D|z|^{\epsilon}}$$
Also, 
$$\begin{align}\prod_{i! > |z|} (1 + \frac{|z|}{i!}) &= e^{\sum_{i! > |z|}(log(1+\frac{|z|}{i!}))} \\
 &\leq e^{\sum_{i! > |z|} (\frac{|z|}{i!})} \\
&= e^{\frac{|z|}{(n+1)!}\sum_{i! > |z|} (\frac{(n+1)!}{i!})} \\
& \leq e^{\frac{2|z|}{n!}} \;\;\; (using \mathbf{(1)}) \\
& \leq e^2
\end{align}$$
If $|z| < 6!$, then $|f(z)|$ is bounded.
Thus, for any $\epsilon >0$, $\;$$\exists \; A, B >0$ $\;$ such that $|f(z)| \leq Ae^{B|z|^\epsilon}$. 
Hence, $p = 0$  
QED
