$$L_{n}\left[\alpha,c\right]:=e^{\left(c + o(1)\right)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}$$
$f$ is in o(1) if and only if for every $c\in\left[0,\infty\right)$, there exists some $N\in\mathbb{N}$ such that for every $n\in\mathbb{N}$ with $n\geq N$, we have $f(n)< c \cdotp 1$.
So, can't we just replace the $\left(c + o(1)\right)$ with just $k$ a new constant? o(1) `vanishes' as $n\rightarrow\infty$ What extra information does$\left(c + o(1)\right)$ add?
Edit to clarify:
$e^{\left(c + o(1)\right)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}=e^{c(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}e^{o(1)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}\rightarrow e^{c(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}$ as $n\rightarrow\infty$
So, $e^{(c+o(1))(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}\leq e^{(c+\epsilon)(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}$ eventually, so let $k=c+\epsilon$, $$L_{n}\left[\alpha,c\right]:=O\left(e^{k(\ln n)^\alpha(\ln \ln n)^{1-\alpha}}\right)$$