The order of an entire function $f$ is defined by the smallest $\alpha$ in $[0,\infty]$ such that $f(z)\ll e^{|z|^{\alpha+\epsilon}}$ for all $\epsilon>0$. An equivalent definition is, if $\limsup_{r\to\infty} {\frac{\log\log M(r,f)}{\log r}}=\lambda$, where $M(r,f)=\max|f(z)|$ on $|z|=r$, then $\lambda$ is the order of $f$.
Question: Let $f(z)=\prod_{n=1}^{\infty}(1+\frac{z}{r_n})^2$ where $0<r_1<r_2<\ldots$. How to show that the order of this entire function is zero?
My attempt: From the first definition, I want to check that whether $\log\log |f(z)|$ grows slower than $\log{|z|^\epsilon}$ or not for all $\epsilon>0$. Now $\log\log |f(z)|=\log(\log|1+\frac{z}{r_1}|^2+ \log|1+\frac{z}{r_2}|^2+\cdots)=\log2+\log(\log|1+\frac{z}{r_1}|+ \log|1+\frac{z}{r_2}|+\cdots)$
From here I am not getting any clue to proceed. From the second definition I am not able to find the function $M(r,f)$.
Any suggestions or comments are welcome.