If $X \sim$ Poisson($\lambda$) and $r = 1, 2, \dotsc$, then I would like to show that $|\mathbb{E}[X^r||^{1/r} = O(r)$.
Attempt: For any $r = 1, 2, \dotsc$, $$ \mathbb{E}[X^r] = \sum_{j=1}^r\lambda^j{r \brace j}, $$ where ${r \brace j}$ is Stirling numbers of the second kind. According to Wikipedia, there exists a positive integer $K_r$ such that for all $1 \leq j \leq r$ and large $r$, $$ {r \brace j} \leq {r \brace K_r} = e^{r\log r- r\log\log r - r + O(r\log\log r/\log r)}. $$ If $\lambda \geq 1$, then $$ |\mathbb{E}[X^r||^{1/r} \leq \left(\sum_{j=1}^r\lambda^r{r \brace K_r}\right)^{1/r} = r^{1/r}\lambda{r \brace K_r}^{1/r} \leq r^2(\log r)^{-1}e^{-1}e^{1/rO(r\log\log r/\log r)} \leq r(\log r)^{-1}e^{-1}e^{1/rO(r\log\log r/\log r)}. $$ For any $r \geq e$, $$ |\mathbb{E}[X^r||^{1/r} \leq r e^{-1}e^{1/rO(r\log\log r/\log r)}. $$
Question: I'm not too familiar with big O-notation, and that's why I'm a bit struggling with understanding $$ e^{1/rO(r\log\log r/\log r)}. $$ Can I say that there exists a constant $M$ such that $$ e^{1/rO(r\log\log r/\log r)} \leq M? $$ If so, why? Because if that is the case, then I believe the proof is finished.