To find the product of the two smallest prime factors of \(2^{1024} - 1\), we start by observing the expression:
\[ 2^{1024} - 1 \]
This expression can be factored using the difference of squares:
\[ 2^{1024} - 1 = (2^{512} - 1)(2^{512} + 1) \]
Continuing to factor \(2^{512} - 1\) using the same difference of squares strategy, we have:
\[ 2^{512} - 1 = (2^{256} - 1)(2^{256} + 1) \]
Proceeding further:
\[ 2^{256} - 1 = (2^{128} - 1)(2^{128} + 1) \]
\[ 2^{128} - 1 = (2^{64} - 1)(2^{64} + 1) \]
\[ 2^{64} - 1 = (2^{32} - 1)(2^{32} + 1) \]
\[ 2^{32} - 1 = (2^{16} - 1)(2^{16} + 1) \]
\[ 2^{16} - 1 = (2^{8} - 1)(2^{8} + 1) \]
\[ 2^{8} - 1 = (2^{4} - 1)(2^{4} + 1) \]
\[ 2^{4} - 1 = (2^{2} - 1)(2^{2} + 1) \]
\[ 2^{2} - 1 = 3 \quad \text{and} \quad 2^{2} + 1 = 5 \]
Thus, we have the factorization:
\[ 2^{4} - 1 = 3 \times 5 \quad \text{and} \quad 2^{4} + 1 = 17 \]
We continue to find that:
\[ 2^{8} - 1 = (2^{4} - 1)(2^{4} + 1) = 15 \times 17 = 3 \times 5 \times 17 \]
And similarly, we keep aggregating the smaller factorizations, leading up to:
\[ 2^{32} - 1 = (2^{16} - 1)(2^{16} + 1) \] continuing this logic leads us to keep identifying the expressions involving 3, 5, 7, 17, etc.
The approach simplifies when we realize \(2^n - 1\) for \(n = 1024, 512, 256,\) etc. has well-known prime factorization patterns. As \(2^n - 1\) contributes repeated patterns of smaller primes and we recognize:
- The smallest prime, \(3\), divides \(2^{2} - 1\).
- The next smallest prime factor present is \(5\).
Now we have confirmed the two smallest prime factors of \(2^{1024} - 1\) are \(3\) and \(5\).
Finally, we calculate the product:
\[ 3 \times 5 = 15 \]
Thus, the product of the two smallest prime factors of \(2^{1024} - 1\) is:
\[ \boxed{15} \]
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