Iteration of $x/\log x$

Question

Consider $f(x)= \frac{x}{\log x}$ iterated n times for given x in $Z^+.$ $f^2 = f \circ f.$

Let $x_1 = x^2.$

What I would like to show (or disprove) : $\exists ~\alpha = x_n - e > 0 $ such that if n is minimal with

$ f^n(x^2) - e < \alpha $ then $$\lim_{x \to \infty} n = \lceil \log x \rceil.$$

Example: Let $\alpha \approx 0.0003$.

$\log(700,000^2) = 13.45 $ and $f^{14}(700,000^2) = 2.7183$ but $f^{13} = 2.727.$

Edit: The motivation may shed a little light on the question. It seems for small numbers that if we have an n with $\pi^n(x^2) = 1,$ for the same n we have $f^n \approx e + \Delta$ for a small (but seemingly non-vanishing) $\Delta >0.$ This, notwithstanding that we are compounding the initial error of the PNT many times over for large $x^2.$ Naively I would expect n to differ for f and $\pi$ as $x^2$ gets large, and maybe that $\Delta$ would go quickly to 0. Instead there seems to be a good correspondence between n for $\pi^n(x^2) $ and $f^n(x^2).$* I don't think this correspondence is tractable but thought maybe something could be said about f.

*The example above is a case in point. $\pi^{14}(700000^2) = 1.$

Answer 1

False. Just try bigger number like $x = 10^{100}$.
$\log(x) \sim 230.258509$ but the corresponding $n$ is 97.
( $f^{97}(x^2) \sim 2.718294$ and $f^{96}(x^2) \sim 2.726563$ ).

EDIT

Let $x_k = f^k(x^2)$, we have $\log(x_{k+1}) = \log(x_{k}) - \log \log(x_{k})$. When $\log(x_k) \gg 1 $, $\log(x_{k}) \gg \log\log(x_{k})$. The limiting behavior of $n$ should not be $\log x$. Instead, should be something like: $$n \sim \int_{?}^{\log x} \frac{dt}{\log t} \sim \frac{\log x}{C(x)}$$ for some function $C(x)$ which diverges to $\infty$ very slowly.