Is it true that $\Gamma^{-1}(x) \in O(\log x/\log \log x)$

Question

Here, $\Gamma^{-1}(x)$ is the inverse of the gamma function. Or, if you prefer, $\Gamma^{-1}(x) = \min \{y : y! \geq x\}$. I saw a claim that $\Gamma^{-1}(x) \in O(\log x/\log\log x)$ but I'm not sure how to show this.

Answer 1

Setting $y = \Gamma^{-1}(x)$, so that $x = \Gamma(y)$, as $x \to \infty$ we have $y \to \infty$. By Stirling's formula we then have

$$ \log x \sim y \log y $$

and, taking logarithms again,

$$ \log\log x \sim \log y $$

as $x \to \infty$. Combining these two formulas yields

$$ \Gamma^{-1}(x) = y \sim \frac{\log x}{\log y} \sim \frac{\log x}{\log\log x}. $$