what is the limit of $\displaystyle (ln^2 k)^{\frac{1}{k}}$
when k goes to infinity
?
$\displaystyle \displaystyle \begin{align*}\lim_{k \to \infty}\left(\ln^2{k}\right)^{\frac{1}{k}} &= \lim_{k \to \infty}e^{\ln{\left[\left(\ln^2{k}\right)^{\frac{1}{k}}\right]}} \\ &= e^{\lim_{k \to \infty}\ln{\left[\left(\ln^2{k}\right)^{\frac{1}{k}}\right]}} \\ &= e^{\lim_{k \to \infty}\left[\frac{\ln{\left(\ln^2{k}\right)}}{k}\right]}\end{align*}$
Now you can apply L'Hospital's Rule.