Let $\mathbb{Z}_{\ge 0}$ denote the set of nonnegative integers.

Define a function $f:\mathbb{Z}_{\ge 0} \to\mathbb{Z}$ with $f\left(0\right)=1$ and \[ f\left(n\right)=512^{\left\lfloor n/10 \right\rfloor}f\left(\left\lfloor n/10 \right\rfloor\right)\]
for all $n \ge 1$. Determine the number of nonnegative integers $n$ such that the hexadecimal (base $16$) representation of $f\left(n\right)$ contains no more than $2500$ digits.

[i]Proposed by Tristan Shin[/i]
1. **Define the function and initial conditions:**
   The function \( f: \mathbb{Z}_{\ge 0} \to \mathbb{Z} \) is defined with \( f(0) = 1 \) and for \( n \ge 1 \),
   \[
   f(n) = 512^{\left\lfloor \frac{n}{10} \right\rfloor} f\left(\left\lfloor \frac{n}{10} \right\rfloor\right).
   \]

2. **Understand the behavior of the function:**
   By examining the recursive definition, we see that \( f(n) \) depends on the value of \( f \) at \( \left\lfloor \frac{n}{10} \right\rfloor \). This implies that \( f(n) \) grows exponentially as \( n \) increases.

3. **Express \( f(n) \) in terms of powers of 512:**
   We can write \( f(n) \) as:
   \[
   f(n) = 512^{\left\lfloor \frac{n}{10} \right\rfloor} f\left(\left\lfloor \frac{n}{10} \right\rfloor\right).
   \]
   By iterating this process, we get:
   \[
   f(n) = 512^{\left\lfloor \frac{n}{10} \right\rfloor + \left\lfloor \frac{\left\lfloor \frac{n}{10} \right\rfloor}{10} \right\rfloor + \left\lfloor \frac{\left\lfloor \frac{\left\lfloor \frac{n}{10} \right\rfloor}{10} \right\rfloor}{10} \right\rfloor + \cdots} f(0).
   \]
   Since \( f(0) = 1 \), we have:
   \[
   f(n) = 512^{\sum_{k=0}^{\infty} \left\lfloor \frac{n}{10^k} \right\rfloor}.
   \]

4. **Simplify the sum:**
   The sum \( \sum_{k=0}^{\infty} \left\lfloor \frac{n}{10^k} \right\rfloor \) is the sum of the digits of \( n \) in base 10, which is approximately \( n \) for large \( n \). Therefore:
   \[
   f(n) \approx 512^n.
   \]

5. **Determine the number of digits in the hexadecimal representation:**
   The number of digits \( d \) in the hexadecimal representation of \( f(n) \) is given by:
   \[
   d = \left\lceil \log_{16}(f(n)) \right\rceil.
   \]
   Since \( f(n) \approx 512^n \), we have:
   \[
   \log_{16}(f(n)) = \log_{16}(512^n) = n \log_{16}(512).
   \]
   Using the change of base formula, \( \log_{16}(512) = \frac{\log_{2}(512)}{\log_{2}(16)} = \frac{9}{4} \).

6. **Set up the inequality for the number of digits:**
   We need the number of digits to be no more than 2500:
   \[
   n \cdot \frac{9}{4} \leq 2500.
   \]
   Solving for \( n \):
   \[
   n \leq \frac{2500 \cdot 4}{9} = \frac{10000}{9} \approx 1111.11.
   \]
   Since \( n \) must be an integer, the largest possible \( n \) is 1111.

7. **Count the number of valid \( n \):**
   The nonnegative integers \( n \) that satisfy this condition are from 0 to 1111, inclusive. Therefore, there are:
   \[
   1111 - 0 + 1 = 1112.
   \]

The final answer is \(\boxed{1112}\).