Palmer correctly computes the product of the first $1,001$ prime numbers. Which of the following is NOT a factor of Palmer's product?

$\text{(A) }2,002\qquad\text{(B) }3,003\qquad\text{(C) }5,005\qquad\text{(D) }6,006\qquad\text{(E) }7,007$
1. Palmer computes the product of the first $1,001$ prime numbers. Let's denote this product as \( P \).
2. We need to determine which of the given numbers is NOT a factor of \( P \). The given numbers are:
   - \( 2,002 \)
   - \( 3,003 \)
   - \( 5,005 \)
   - \( 6,006 \)
   - \( 7,007 \)

3. First, let's factorize each of these numbers into their prime factors:
   - \( 2,002 = 2 \times 7 \times 11 \times 13 \)
   - \( 3,003 = 3 \times 7 \times 11 \times 13 \)
   - \( 5,005 = 5 \times 7 \times 11 \times 13 \)
   - \( 6,006 = 2 \times 3 \times 7 \times 11 \times 13 \)
   - \( 7,007 = 7^2 \times 11 \times 13 \)

4. Since \( P \) is the product of the first $1,001$ prime numbers, it includes each of the first $1,001$ primes exactly once.

5. Now, let's check each number to see if it can be a factor of \( P \):
   - \( 2,002 = 2 \times 7 \times 11 \times 13 \): All these primes (2, 7, 11, 13) are among the first $1,001$ primes, so \( 2,002 \) is a factor of \( P \).
   - \( 3,003 = 3 \times 7 \times 11 \times 13 \): All these primes (3, 7, 11, 13) are among the first $1,001$ primes, so \( 3,003 \) is a factor of \( P \).
   - \( 5,005 = 5 \times 7 \times 11 \times 13 \): All these primes (5, 7, 11, 13) are among the first $1,001$ primes, so \( 5,005 \) is a factor of \( P \).
   - \( 6,006 = 2 \times 3 \times 7 \times 11 \times 13 \): All these primes (2, 3, 7, 11, 13) are among the first $1,001$ primes, so \( 6,006 \) is a factor of \( P \).
   - \( 7,007 = 7^2 \times 11 \times 13 \): The prime 7 appears twice in the factorization of \( 7,007 \). Since \( P \) includes each prime exactly once, \( 7,007 \) cannot be a factor of \( P \).

Conclusion:
The number that is NOT a factor of Palmer's product is \( 7,007 \).

The final answer is \( \boxed{7,007} \).