Let $\{n,p\}\in\mathbb{N}\cup \{0\}$ such that $2p\le n$. Prove that $\frac{(n-p)!}{p!}\le \left(\frac{n+1}{2}\right)^{n-2p}$. Determine all conditions under which equality holds.
1. **Initial Assumptions and Simplifications:**
   - Given \( \{n, p\} \in \mathbb{N} \cup \{0\} \) such that \( 2p \le n \).
   - We need to prove \( \frac{(n-p)!}{p!} \le \left( \frac{n+1}{2} \right)^{n-2p} \).
   - We first consider the case \( n = 2p \). Here, \( \frac{(n-p)!}{p!} = \frac{(2p-p)!}{p!} = \frac{p!}{p!} = 1 \) and \( \left( \frac{n+1}{2} \right)^{n-2p} = \left( \frac{2p+1}{2} \right)^0 = 1 \). Thus, the equality holds.
   - Now, assume \( n > 2p \).

2. **Applying AM-GM Inequality:**
   - By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, for any non-negative real numbers \( a_1, a_2, \ldots, a_k \), we have:
     \[
     \sqrt[k]{a_1 a_2 \cdots a_k} \le \frac{a_1 + a_2 + \cdots + a_k}{k}
     \]
   - Let \( k = n - 2p \) and consider the sequence \( a_i = \frac{(n-p)!}{p!} \) for \( i = 1, 2, \ldots, k \).

3. **Calculating the Arithmetic Mean:**
   - The arithmetic mean of the sequence \( a_i \) is:
     \[
     \frac{a_1 + a_2 + \cdots + a_k}{k} = \frac{(n-p) + (n-p-1) + \cdots + (p+1)}{n-2p}
     \]
   - This simplifies to:
     \[
     \frac{(n-p) + (n-p-1) + \cdots + (p+1)}{n-2p} = \frac{(n-p) + (n-p-1) + \cdots + (p+1)}{n-2p}
     \]
   - The sum of the first \( n-2p \) terms of an arithmetic sequence starting from \( n-p \) and ending at \( p+1 \) is:
     \[
     \frac{(n-p) + (p+1)}{2} = \frac{n+1}{2}
     \]

4. **Applying AM-GM to the Sequence:**
   - By AM-GM, we have:
     \[
     \sqrt[n-2p]{\frac{(n-p)!}{p!}} \le \frac{n+1}{2}
     \]
   - Raising both sides to the power \( n-2p \), we get:
     \[
     \frac{(n-p)!}{p!} \le \left( \frac{n+1}{2} \right)^{n-2p}
     \]

5. **Equality Condition:**
   - The equality condition in AM-GM holds if and only if all the numbers in the sequence are equal.
   - This implies that \( n = 2p \) or \( n = 2p + 1 \).

The final answer is \(\boxed{\frac{(n-p)!}{p!} \le \left( \frac{n+1}{2} \right)^{n-2p}}\)