How can I prove, that $$\log^x{x} > x^{\sqrt{x}}$$ for big $n$ ? I tried to logarithm those expressions, deduct them, somehow estimate the values but no luck.
After few tries, I ended up with expression, where I need to prove that $\log{n}$ grows faster than $$\large{n^{\frac{1}{\sqrt{n}}}}$$, so that is an alternative question.
We have $$\log(x)^x = \exp\left(\log\left(\log(x)^x\right)\right) = \exp\left(x\log(\log(x))\right)$$ and $$x^{\sqrt{x}} = \exp\left(\log\left(x^{\sqrt{x}}\right)\right) = \exp\left(\sqrt{x}\log(x)\right)$$ Now conclude what you want by noticing that $$\sqrt{x}\log(x) \ll x \ll x\log(\log(x))$$
The first part of the inequality is true, since $\log(y) \ll y \implies \log(\sqrt{y}) \ll \sqrt{y} \implies \log(y) \ll \sqrt{y}$