A number theory problem by Mateus Gomes

Putting $n=10$

$3^{2^{10}}-1=(3^{2^9}+1)(3^{2^9}-1)\\ =(3^{2^9}+1)(3^{2^8}+1)(3^{2^8}-1) \\ = \cdots \\ =(3^{2^9}+1)(3^{2^8}+1)(3^{2^8}-1)...(3^{2^2}+1)(3^{2}+1)(3+1)(3-1)$

Each of the 12 factors will release exactly a single factor of 2 because in the binomial expansion of $3^{2 k}+1$ will give a factor 2 and rest part is odd.

So the number will be $3^{2^{10}}-1=2^{12}(2 m+1) \\ =2^{12}.2m +2^{12} \\ =2^{13}m + 2^{12}$

So the remainder is $2^{12}=4096$