Need explanation on a problem involving natural function Consider a positive integer $n$ and the function
$f:\mathbb{N}\to \mathbb{N}$ ($\mathbb N$ includes $0$) by
$$f(x) = \begin{cases}   \frac{x}{2}
 & \text{if } x \text{ is even} \\   \frac{x-1}{2} + 2^{n-1} & \text{if } x \text{ is odd} \end{cases} $$
Determine the set
$$   A = \{ x\in \mathbb{N} \mid \underbrace{\left( f\circ f\circ ....\circ f \right)}_{n\ f\text{'s}}\left( x \right)=x \}.   $$
(Romania NMO 2013)
The solution starts by stating that $f(x)<x, \quad\forall x\ge 2^n-1$. This was easy enough to understand. However, they continue by saying this implies that $A\subset\{0,1,\dots,2^n-1\}$. Why is that?
Please help me understand! Thanks in advance!

Hint:
Letting all divisions be integer and splitting the cases after the low order bits of $x$, we can expand the function iterates for, say, $n=4$, and the pattern becomes obvious.
$$f(x)=\begin{cases}\frac x2&\text{ if } x \text{ is even}\\\frac{x}2+8&\text{ if } x \text{ is odd}\end{cases}$$
$$f(f(x))=\begin{cases}\frac x4&\\\frac{x}4+4\\\frac{x}4+8\\\frac{x}4+12\end{cases}$$
$$f(f(f(x)))=\begin{cases}
\frac x8&\\\frac{x}8+2\\\frac{x}8+4\\\frac{x}8+6\\
\frac x8+8&\\\frac{x}8+10\\\frac{x}8+12\\\frac{x}8+14\end{cases}$$
$$f(f(f(f(x))))=\begin{cases}
\frac x{16}&\\\frac{x}{16}+1\\\frac{x}{16}+2\\\frac{x}{16}+3\\
\frac x{16}+4&\\\frac{x}{16}+5\\\frac{x}{16}+6\\\frac{x}{16}+7\\
\frac x{16}+8&\\\frac{x}{16}+9\\\frac{x}{16}+10\\\frac{x}{16}+11\\
\frac x{16}+12&\\\frac{x}{16}+13\\\frac{x}{16}+14\\\frac{x}{16}+15\end{cases}$$
