Which of the following polynomials does not divide $x^{60} - 1$?

$ 
\textbf{a)}\ x^2+x+1
\qquad\textbf{b)}\ x^4-1
\qquad\textbf{c)}\ x^5-1
\qquad\textbf{d)}\ x^{15}-1
\qquad\textbf{e)}\ \text{None of above}
$
To determine which of the given polynomials does not divide \( x^{60} - 1 \), we will analyze each polynomial individually.

1. **Analyzing \( x^2 + x + 1 \):**
   - The roots of \( x^2 + x + 1 \) are the non-real cube roots of unity, denoted by \( \omega \) and \( \omega^2 \), where \( \omega \) is a primitive cube root of unity. This means \( \omega^3 = 1 \) and \( \omega \neq 1 \).
   - Since \( \omega^3 = 1 \), we have \( \omega^{60} = (\omega^3)^{20} = 1^{20} = 1 \). Therefore, \( \omega \) and \( \omega^2 \) are roots of \( x^{60} - 1 \).
   - Hence, \( x^2 + x + 1 \) divides \( x^{60} - 1 \).

2. **Analyzing \( x^4 - 1 \):**
   - We can factor \( x^4 - 1 \) as \( (x^2 - 1)(x^2 + 1) \).
   - Further factorization gives \( (x - 1)(x + 1)(x^2 + 1) \).
   - Each of these factors divides \( x^{60} - 1 \) because \( x^{60} - 1 \) includes all roots of unity of order dividing 60.
   - Therefore, \( x^4 - 1 \) divides \( x^{60} - 1 \).

3. **Analyzing \( x^5 - 1 \):**
   - The polynomial \( x^5 - 1 \) has roots that are the 5th roots of unity.
   - Since \( 5 \mid 60 \), \( x^{60} - 1 \) includes all 5th roots of unity.
   - Therefore, \( x^5 - 1 \) divides \( x^{60} - 1 \).

4. **Analyzing \( x^{15} - 1 \):**
   - The polynomial \( x^{15} - 1 \) has roots that are the 15th roots of unity.
   - Since \( 15 \mid 60 \), \( x^{60} - 1 \) includes all 15th roots of unity.
   - Therefore, \( x^{15} - 1 \) divides \( x^{60} - 1 \).

Since all the given polynomials \( x^2 + x + 1 \), \( x^4 - 1 \), \( x^5 - 1 \), and \( x^{15} - 1 \) divide \( x^{60} - 1 \), the correct answer is:

\(\boxed{\text{e}}\) None of the above.