Let $S$ be a set. We say $S$ is $D^\ast$[i]-finite[/i] if there exists a function $f : S \to S$ such that for every nonempty proper subset $Y \subsetneq S$, there exists a $y \in Y$ such that $f(y) \notin Y$.  The function $f$ is called a [i]witness[/i] of $S$.  How many witnesses does $\{0,1,\cdots,5\}$ have?

[i]Proposed by Evan Chen[/i]
To determine the number of witnesses for the set \( \{0, 1, \cdots, 5\} \), we need to understand the definition of a witness function \( f \) for a set \( S \). A function \( f: S \to S \) is a witness if for every nonempty proper subset \( Y \subsetneq S \), there exists a \( y \in Y \) such that \( f(y) \notin Y \).

1. **Base Case:**
   Consider the smallest non-trivial set, \( S = \{0\} \). For this set, there are no nonempty proper subsets, so any function \( f: \{0\} \to \{0\} \) trivially satisfies the condition. Thus, there is exactly one witness function for \( \{0\} \).

2. **Inductive Hypothesis:**
   Assume that for a set \( \{0, 1, \cdots, n-1\} \), the number of witness functions is \( (n-1)! \).

3. **Inductive Step:**
   We need to show that for the set \( \{0, 1, \cdots, n\} \), the number of witness functions is \( n! \).

   - Consider any permutation \( \sigma \) of the set \( \{0, 1, \cdots, n\} \). Define \( f \) such that \( f(i) = \sigma(i) \) for all \( i \in \{0, 1, \cdots, n\} \).
   - We need to verify that \( f \) is a witness function. Take any nonempty proper subset \( Y \subsetneq \{0, 1, \cdots, n\} \). Since \( Y \) is a proper subset, there exists at least one element \( k \in \{0, 1, \cdots, n\} \) that is not in \( Y \). The permutation \( \sigma \) ensures that \( \sigma \) maps elements to distinct elements, so there must be some \( y \in Y \) such that \( \sigma(y) \notin Y \). Hence, \( f(y) = \sigma(y) \notin Y \).

   - Therefore, every permutation of \( \{0, 1, \cdots, n\} \) is a valid witness function. The number of such permutations is \( n! \).

4. **Conclusion:**
   By induction, the number of witness functions for the set \( \{0, 1, \cdots, n\} \) is \( n! \).

For the set \( \{0, 1, \cdots, 5\} \), the number of witness functions is:
\[ 5! = 120 \]

The final answer is \( \boxed{120} \).