6. (10 points) Try to determine the last two digits of the product $\left(2^{1}+1\right)\left(2^{2}+1\right)\left(2^{3}+1\right) \cdots\left(2^{2011}+1\right)$.
【Analysis】First, determine that the product can be divisible by 25. Since each factor is odd, the last two digits of the product are either 25 or 75. By examining the remainders of each factor when divided by 4, we can determine the remainder of the product, and thus conclude that the last two digits are 75.

【Solution】Solution: Let \( n = (2^1 + 1) \times (2^2 + 1) \times (2^3 + 1) \times \cdots \times (2^{2011} + 1) \),

Since each factor \( 2^k + 1 \) is odd, and \( 2^2 + 1 = 5 \), \( 2^6 + 1 = 65 = 5 \times 13 \), we have \( n \equiv 0 \pmod{25} \). This means the last two digits of \( n \) are either 25 or 75.

Also, \( 2^1 + 1 \equiv 3 \pmod{4} \), and for \( k \geq 2 \), \( 2^k + 1 \equiv 1 \pmod{4} \). Therefore, \( n \equiv 3 \pmod{4} \), meaning the last two digits of \( n \) leave a remainder of 3 when divided by 4. Since \( 25 \equiv 1 \pmod{4} \) and \( 75 \equiv 3 \pmod{4} \), the last two digits of \( n \) are 75.

Answer: The last two digits of \( (2^1 + 1)(2^2 + 1)(2^3 + 1) \cdots (2^{2011} + 1) \) are 75.