Find the largest integer $n$ such that $2007^{1024}-1$ is divisible by $2^n$.

$\textbf{(A) }1\hspace{14em}\textbf{(B) }2\hspace{14em}\textbf{(C) }3$

$\textbf{(D) }4\hspace{14em}\textbf{(E) }5\hspace{14em}\textbf{(F) }6$

$\textbf{(G) }7\hspace{14em}\textbf{(H) }8\hspace{14em}\textbf{(I) }9$

$\textbf{(J) }10\hspace{13.7em}\textbf{(K) }11\hspace{13.5em}\textbf{(L) }12$

$\textbf{(M) }13\hspace{13.3em}\textbf{(N) }14\hspace{13.4em}\textbf{(O) }15$

$\textbf{(P) }16\hspace{13.6em}\textbf{(Q) }55\hspace{13.4em}\textbf{(R) }63$

$\textbf{(S) }64\hspace{13.7em}\textbf{(T) }2007$
To find the largest integer \( n \) such that \( 2007^{1024} - 1 \) is divisible by \( 2^n \), we need to determine the highest power of 2 that divides \( 2007^{1024} - 1 \).

1. **Factorization**:
   We start by factoring \( 2007^{1024} - 1 \) using the difference of powers:
   \[
   2007^{1024} - 1 = (2007^{512} - 1)(2007^{512} + 1)
   \]
   We can continue factoring each term:
   \[
   2007^{512} - 1 = (2007^{256} - 1)(2007^{256} + 1)
   \]
   \[
   2007^{256} - 1 = (2007^{128} - 1)(2007^{128} + 1)
   \]
   \[
   \vdots
   \]
   \[
   2007^2 - 1 = (2007 - 1)(2007 + 1)
   \]
   \[
   2007 - 1 = 2006 \quad \text{and} \quad 2007 + 1 = 2008
   \]

2. **Analyzing Factors**:
   We need to determine the number of factors of 2 in each term. Let's start with \( 2006 \) and \( 2008 \):
   \[
   2006 = 2 \times 1003 \quad \text{(one factor of 2)}
   \]
   \[
   2008 = 2^3 \times 251 \quad \text{(three factors of 2)}
   \]

3. **Analyzing Remaining Factors**:
   For the remaining factors of the form \( 2007^{2^k} + 1 \):
   \[
   2007 \equiv -1 \pmod{4} \implies 2007^{2^k} \equiv 1 \pmod{4}
   \]
   \[
   2007^{2^k} + 1 \equiv 1 + 1 = 2 \pmod{4}
   \]
   Each of these terms contributes exactly one factor of 2.

4. **Counting Factors**:
   There are 10 such terms (from \( 2007^2 + 1 \) to \( 2007^{1024} + 1 \)), each contributing one factor of 2:
   \[
   10 \text{ factors of 2 from } 2007^{2^k} + 1
   \]
   Adding the factors from \( 2006 \) and \( 2008 \):
   \[
   1 \text{ (from 2006)} + 3 \text{ (from 2008)} + 10 \text{ (from other terms)} = 14
   \]

Thus, the largest power of 2 that divides \( 2007^{1024} - 1 \) is \( 2^{14} \).

The final answer is \( \boxed{14} \)