Initially, I need to prove, that
$$\forall \lambda > 0 \ \forall\, \xi_i \sim \text{Pois}(\lambda), \xi_i\text{ are independent } \implies P\left(\limsup\limits_{n \rightarrow \infty}\frac{\xi_n \log{\log{n}}}{\log{n}} = 1\right) = 1$$
I used Borel-Cantelli lemma and Taylor series with Lagrange remainder, and reduced problem to the following:
$\forall a > 0, \ \forall\, \varepsilon > 0 \ \sum_{n=2}^{\infty}{(a\ \log(n)/\log(\log(n)))^{-(1-\varepsilon)\log(n)/\log(\log(n)}}$ - diverges and
$\sum_{n=2}^{\infty}{(a\log(n)/\log(\log(n)))^{-(1+\varepsilon)\log(n)/\log(\log(n)}}$ converges.
WolframAlpha tells me that's true, but I can't prove it...