Let $x$ and $n$ be integers such that $1\le x\le n$. We have $x+1$ separate boxes and $n-x$ identical balls. Define $f(n,x)$ as the number of ways that the $n-x$ balls can be distributed into the $x+1$ boxes. Let $p$ be a prime number. Find the integers $n$ greater than $1$ such that the prime number $p$ is a divisor of $f(n,x)$ for all $x\in\{1,2,\ldots ,n-1\}$.
1. We start by understanding the function \( f(n, x) \). The problem states that \( f(n, x) \) is the number of ways to distribute \( n - x \) identical balls into \( x + 1 \) separate boxes. This is a classic combinatorial problem that can be solved using the "stars and bars" theorem. According to this theorem, the number of ways to distribute \( k \) identical items into \( r \) distinct groups is given by:
   \[
   \binom{k + r - 1}{r - 1}
   \]
   Here, \( k = n - x \) and \( r = x + 1 \). Therefore, we have:
   \[
   f(n, x) = \binom{(n - x) + (x + 1) - 1}{(x + 1) - 1} = \binom{n}{x}
   \]

2. Next, we need to find the integers \( n \) greater than 1 such that the prime number \( p \) is a divisor of \( f(n, x) = \binom{n}{x} \) for all \( x \in \{1, 2, \ldots, n-1\} \).

3. According to Lucas' theorem, for a prime \( p \), the binomial coefficient \( \binom{n}{x} \) modulo \( p \) can be computed using the base \( p \) representations of \( n \) and \( x \). Specifically, if \( n = n_k n_{k-1} \ldots n_0 \) and \( x = x_k x_{k-1} \ldots x_0 \) are the base \( p \) representations of \( n \) and \( x \), then:
   \[
   \binom{n}{x} \equiv \prod_{i=0}^{k} \binom{n_i}{x_i} \pmod{p}
   \]
   For \( \binom{n}{x} \) to be divisible by \( p \) for all \( x \in \{1, 2, \ldots, n-1\} \), it must be that at least one of the binomial coefficients \( \binom{n_i}{x_i} \) is zero modulo \( p \) for all \( x \).

4. The binomial coefficient \( \binom{n_i}{x_i} \) is zero modulo \( p \) if and only if \( n_i < x_i \). For this to hold for all \( x \in \{1, 2, \ldots, n-1\} \), it must be that \( n \) is a power of \( p \). That is, \( n = p^m \) for some \( m \in \mathbb{N} \).

5. Therefore, the integers \( n \) greater than 1 such that \( p \) is a divisor of \( f(n, x) \) for all \( x \in \{1, 2, \ldots, n-1\} \) are precisely the powers of \( p \).

The final answer is \( \boxed{ n = p^m } \) for some \( m \in \mathbb{N} \).