Maximum power of $2$ which divides $3^{1024}-1$ What is the maximum power of $2$ which completely divides $3^{1024}-1$?
I proceeded thus:
$\phi(2^n)=2^{n-1}$ for all $n\ge1$
$$3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}$$
$$3^{1024}-1\equiv0\pmod {2^{11}}$$
Since $\phi(2^{11})=2^{10}$.  So, maximum power of $2$ must be $11$.
But the answer says it is $12$.
Where am I wrong and how to solve it correctly? 

Just to give a different approach, since $3^4=81=1+80=1+5\cdot2^4$, we have
$$3^{1024}-1=(1+80)^{256}-1=256\cdot80+{256\choose2}80^2+{256\choose3}80^3+\cdots\\
=2^8(5\cdot2^4)+(2^7\cdot255)(5^2\cdot2^8)+(2^7\cdot85\cdot254)(5^3\cdot2^{12})+\cdots\\
\equiv5\cdot2^{12}\mod2^{13}$$
all terms after the first having higher and higher powers of $2$. Thus $3^{1024}-1$ is divisible by $2^{12}$ but not by $2^{13}$.
