Induction on a piecewise defined function Let $f \colon [ 0,\infty) \to [0,\infty)$ with $f(x):= \begin{cases}  x, & x \leq 1, \\
     f(\sqrt{x-1}\,), & x > 1. \end{cases}$
I made some plots of this function and I am pretty aware that it is bounded (even by very sloppy reasons: well, $\sqrt{x-1} < x$ for $x > 1$ and after a while, we will be in the base-case, which is bounded by $1$.)
I am looking for a nice induction on this fact. But I am struggling to find a reasonable formulation for that. 
I want to proof boundedness. Is it a clever way to use induction on $n \in \mathbb{N}$ to show that for any $x \in \left[0, n \right]$, $f(x)$ is bounded by 1?
Hypothesis: $f(x)$ is bounded on $[0, \infty)$ by $1$.
Base case: $f(x)$ is bounded by $1$ on $[0, 1)$: by defintion.
Induction: If $x \in [0, n )$, there is nothing to prove. For $x \in  \left[n, n+1 \right)$, we have $f(x) = f(\sqrt{x-1})$. Because $x \in [n, n+1)$, $x-1 \in [n-1, n) \subseteq [0, n)$ and the induction hypothesis can be used. 
Is this reasoning valid or do I need to have a double-base-case because of $[n-1, n)$.
 A: The function as defined by OP never maps to any number outside the interval $[0,1]$.
If we define a sequence $a_0=0$, $a_{n+1}=a_n^2+1$ the first few terms of which are 
$$ 0,1,2,5,26,677,458330,\ldots$$
then we may divide the interval $(0,\infty)$ into the union of disjoint intervals
$$ (0,1],(1,2],(2,5],(5,26],(26,677),\ldots,(a_k,a_{k+1}],\ldots$$
Consider the following sequence for $x>1$:
For any $x=x_0\in(1,\infty)$ define the finite decreasing sequence
$$ x_0,x_1,x_2,\ldots ,x_N=x,\sqrt{x-1},\sqrt{\sqrt{x-1}-1},\ldots x_N\in(0,1] $$
Suppose $x>1$ and $a_N<x\le a_{N+1}$.
Then $a_N<x\le a_N^2+1$ so $a_N-1<x-1\le a_N^2$. 
So $\sqrt{a_N-1}<\sqrt{x-1}\le a_N$. 
But $\sqrt{a_N-1}=a_{N-1}$. 
Therefore, if $a_N<x\le a_{N+1}$ then $a_{N-1}<\sqrt{x-1}\le a_N$. 
But $x_1=\sqrt{x-1}$.
Therefore with $x=x_0>1$
\begin{eqnarray}
x_0&\in&\left(a_{N},a_{N+1}\right]\\
x_1&\in&\left(a_{N-1},a_N\right]\\
x_2&\in&\left(a_{N-2},a_{N-1}\right]\\
& \vdots\\
x_N&\in&(a_0,a_1]=(0,1]
\end{eqnarray}
Therefore, for $x=x_0>1$ 
$$ f(x_0)=f(x_1)=f(x_2)=f(x_3)=\cdots=f(x_N)=x_N\le1$$
Here is a graph of $f$ on the interval $[0,26]$

