Question 10
Let
$$P=(2024^2+1)(2024^{2^2}+1)(2024^{2^3}+1)\cdots(2024^{2^{10}}+1)\times2025+\frac1{2023}.$$
Find the smallest positive integer $N$ such that $N>\log_{2024}P$.
$$P=(2024^2+1)(2024^{2^2}+1)(2024^{2^3}+1)\cdots(2024^{2^{10}}+1)\times2025+\frac1{2023}.$$
Find the smallest positive integer $N$ such that $N>\log_{2024}P$.