Let $p_n$ be the $n^{\mbox{th}}$ prime counting from the smallest prime $2$ in increasing order. For example, $p_1=2, p_2=3, p_3 =5, \cdots$

(a) For a given $n \ge 10$, let $r$ be the smallest integer satisfying 
\[2\le r \le n-2, \quad n-r+1 < p_r\]
and define $N_s=(sp_1p_2\cdots p_{r-1})-1$ for $s=1,2,\ldots, p_r$. Prove that there exists $j, 1\le j \le p_r$, such that none of $p_1,p_2,\cdots, p_n$ divides $N_j$.

(b) Using the result of (a), find all positive integers $m$ for which 
\[p_{m+1}^2 < p_1p_2\cdots p_m\]
### Part (a)

1. **Define the sequence and the integer \( r \):**
   - Let \( p_n \) be the \( n \)-th prime number.
   - For a given \( n \ge 10 \), let \( r \) be the smallest integer satisfying:
     \[
     2 \le r \le n-2, \quad n-r+1 < p_r
     \]

2. **Define \( N_s \):**
   - Define \( N_s = (sp_1p_2\cdots p_{r-1}) - 1 \) for \( s = 1, 2, \ldots, p_r \).

3. **Show that there exists \( j \) such that none of \( p_1, p_2, \ldots, p_n \) divides \( N_j \):**
   - Note that \( p_1, p_2, \ldots, p_{r-1} \) do not divide any \( N_s \) because \( N_s \equiv -1 \pmod{p_i} \) for \( i = 1, 2, \ldots, r-1 \).
   - Each \( N_s \) must have a prime factor distinct from \( p_1, p_2, \ldots, p_{r-1} \).
   - Each of \( p_r, p_{r+1}, \ldots, p_n \) can divide at most one of the \( N_s \).
   - There are \( p_r \) values of \( N_s \), and since \( p_r > n-r+1 \), there are more \( N_s \) than primes \( p_r, p_{r+1}, \ldots, p_n \).
   - Therefore, there must be at least one \( N_j \) which is not divisible by any of \( p_1, p_2, \ldots, p_n \).

\(\blacksquare\)

### Part (b)

1. **Using the result of (a), find all positive integers \( m \) for which \( p_{m+1}^2 < p_1p_2\cdots p_m \):**
   - From part (a), we know there exists \( 1 \le k \le p_r \) such that \( p_i \nmid N_k \) for \( i = 1, 2, \ldots, n \).
   - This implies \( p_{n+1} \le N_k < N_k + 1 = kp_1p_2\cdots p_{r-1} < p_1p_2\cdots p_r \).

2. **Consider Bonse's Inequality:**
   - Bonse's Inequality states that for \( m \ge 4 \), \( p_{m+1}^2 < p_1p_2\cdots p_m \).

3. **Analyze the inequality:**
   - Given \( r \ge 4 \), we have \( n - (r-1) + 1 \ge p_{r-1} \ge r + 2 \implies n \ge 2r \).
   - Therefore:
     \[
     p_{n+1}^2 < (p_1p_2\cdots p_r)(p_1p_2\cdots p_r) < (p_1p_2\cdots p_r)(p_{r+1}\cdots p_{2r}) < p_1p_2\cdots p_n
     \]
   - The last step holds because \( n \ge 2r \).

4. **Conclusion:**
   - For \( m \ge 4 \), Bonse's Inequality holds, and thus \( p_{m+1}^2 < p_1p_2\cdots p_m \).

\(\blacksquare\)

The final answer is \( \boxed{ m \ge 4 } \).