If $f(x) = x(1 - x^n)^{-1/n}$, how many function compositions does it take for $f(f(f(\cdots f(x_0)))) > 1$?

Question

Fix an integer $n \geq 1$ and let $f(x) := x(1 - x^n)^{-1/n}$. Fix $x_0 \in (0, 1)$. Then there exists an $N$ (depending on $n$ and $x_0$) such that $f(f(f(\cdots f(x_0)))) > 1$ where the function composition occurred $N$ times. Is there a way to estimate the order of magnitude of $N$ depending on $n$ and $x_0$, like what does $N$ asymptotically look like?

Answer 1

For $k \in \mathbb{Z}_+$:

$$\underbrace{f(\cdots f}_{k}(x_0)\cdots) = x (1 - k x_0^n)^{(-1/n)}$$

So just set this function to $1$ and solve for $k$:

$$k = \lceil (-1 + (1/x_0)^n) (1/x_0)^{-n} x_0^{-n} \rceil$$