
	 Determine the number of integers $2 \le n \le 2016$ such that $n^n-1$ is divisible by $2$, $3$, $5$, $7$.
To determine the number of integers \(2 \le n \le 2016\) such that \(n^n - 1\) is divisible by \(2\), \(3\), \(5\), and \(7\), we need to analyze the conditions under which \(n^n - 1\) is divisible by these primes.

1. **Divisibility by 2:**
   - For \(n^n - 1\) to be divisible by 2, \(n\) must be odd. This is because if \(n\) is even, \(n^n\) is even, and \(n^n - 1\) is odd, which is not divisible by 2.

2. **Divisibility by 3:**
   - For \(n^n - 1\) to be divisible by 3, \(n^n \equiv 1 \pmod{3}\). We need to find the values of \(n\) modulo 3 that satisfy this condition.
   - If \(n \equiv 0 \pmod{3}\), then \(n^n \equiv 0 \pmod{3}\), which does not satisfy the condition.
   - If \(n \equiv 1 \pmod{3}\), then \(n^n \equiv 1 \pmod{3}\), which satisfies the condition.
   - If \(n \equiv 2 \pmod{3}\), then \(n^n \equiv 2^n \pmod{3}\). Since \(2^2 \equiv 1 \pmod{3}\), \(2^n \equiv 1 \pmod{3}\) if \(n\) is even, but \(n\) must be odd. Therefore, \(n \equiv 1 \pmod{3}\).

3. **Divisibility by 5:**
   - For \(n^n - 1\) to be divisible by 5, \(n^n \equiv 1 \pmod{5}\). We need to find the values of \(n\) modulo 5 that satisfy this condition.
   - If \(n \equiv 0 \pmod{5}\), then \(n^n \equiv 0 \pmod{5}\), which does not satisfy the condition.
   - If \(n \equiv 1 \pmod{5}\), then \(n^n \equiv 1 \pmod{5}\), which satisfies the condition.
   - If \(n \equiv 2 \pmod{5}\), then \(n^n \equiv 2^n \pmod{5}\). Since \(2^4 \equiv 1 \pmod{5}\), \(2^n \equiv 1 \pmod{5}\) if \(n\) is a multiple of 4, but \(n\) must be odd. Therefore, \(n \equiv 1 \pmod{5}\).

4. **Divisibility by 7:**
   - For \(n^n - 1\) to be divisible by 7, \(n^n \equiv 1 \pmod{7}\). We need to find the values of \(n\) modulo 7 that satisfy this condition.
   - If \(n \equiv 0 \pmod{7}\), then \(n^n \equiv 0 \pmod{7}\), which does not satisfy the condition.
   - If \(n \equiv 1 \pmod{7}\), then \(n^n \equiv 1 \pmod{7}\), which satisfies the condition.
   - If \(n \equiv 2 \pmod{7}\), then \(n^n \equiv 2^n \pmod{7}\). Since \(2^3 \equiv 1 \pmod{7}\), \(2^n \equiv 1 \pmod{7}\) if \(n\) is a multiple of 3, but \(n\) must be odd. Therefore, \(n \equiv 1 \pmod{7}\).

Combining all these conditions, we have:
\[ n \equiv 1 \pmod{210} \]

This means \(n = 210k + 1\) for some integer \(k\). We need to find the values of \(k\) such that \(2 \le n \le 2016\):
\[ 2 \le 210k + 1 \le 2016 \]
\[ 1 \le 210k \le 2015 \]
\[ \frac{1}{210} \le k \le \frac{2015}{210} \]
\[ 1 \le k \le 9.59 \]

Since \(k\) must be an integer, \(k\) can take values from 1 to 9. Therefore, there are 9 possible values of \(n\).

The final answer is \(\boxed{9}\).