If $2^n$ divides $5^{256} - 1$, what is the largest possible value of $n$?

$ 
\textbf{a)}\ 8
\qquad\textbf{b)}\ 10
\qquad\textbf{c)}\ 11
\qquad\textbf{d)}\ 12
\qquad\textbf{e)}\ \text{None of above}
$
To determine the largest possible value of \( n \) such that \( 2^n \) divides \( 5^{256} - 1 \), we need to analyze the factorization of \( 5^{256} - 1 \) and count the powers of 2 in the factorization.

1. **Factorization using difference of powers:**
   \[
   5^{256} - 1 = (5 - 1)(5 + 1)(5^2 + 1)(5^4 + 1)(5^8 + 1)(5^{16} + 1)(5^{32} + 1)(5^{64} + 1)(5^{128} + 1)
   \]
   This factorization is derived from the identity for the difference of powers:
   \[
   a^n - 1 = (a - 1)(a + 1)(a^2 + 1)(a^4 + 1) \cdots (a^{2^{k-1}} + 1)
   \]
   where \( n = 2^k \).

2. **Analyzing each factor modulo 4:**
   - \( 5 \equiv 1 \pmod{4} \)
   - \( 5^2 \equiv 1 \pmod{4} \)
   - \( 5^4 \equiv 1 \pmod{4} \)
   - Generally, \( 5^{2^i} \equiv 1 \pmod{4} \)

   Therefore, for each \( i \):
   \[
   5^{2^i} + 1 \equiv 2 \pmod{4}
   \]
   This implies that each term \( 5^{2^i} + 1 \) is of the form \( 4k + 2 \), which is divisible by 2 but not by 4.

3. **Counting the powers of 2:**
   - The first term \( (5 - 1) = 4 \) contributes \( 2^2 \).
   - The second term \( (5 + 1) = 6 \) contributes \( 2^1 \).
   - Each subsequent term \( 5^{2^i} + 1 \) for \( i = 1, 2, \ldots, 7 \) contributes \( 2^1 \).

   Therefore, the total power of 2 in the factorization is:
   \[
   2^2 \cdot 2^1 \cdot 2^1 \cdot 2^1 \cdot 2^1 \cdot 2^1 \cdot 2^1 \cdot 2^1 \cdot 2^1 = 2^{2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1} = 2^{10}
   \]

Thus, the largest possible value of \( n \) such that \( 2^n \) divides \( 5^{256} - 1 \) is \( 10 \).

The final answer is \( \boxed{10} \).