Find the maximum integer value $n$ such that $2^n\mid 3^{1024} -1$ Since $2^{10} = 1024$ then we can $3^{1024}-1 = ( 3 - 1 )\Pi_{i=0}^{9} {3^{2^i}+1}$
And then we can start eliminating the $3-1=2$ and $(3^2-1)= 2\times 5$
But then? I guess I could calculate, but this is not different as starting calculating in first place.
What is a better method?

Hint $1$:
Note that $1024=2^{10}$
$$3^{2^{10}}-1=(2+1)^{2^{10}}-1=\sum_{k=0}^{2^{10}-1}\dbinom{2^{10}}{k}$$
Hint $2$:
$3^{1024}-1$ is the difference of squares. Consider the factors to get the answer.
