Showing that $2^6$ divides $3^{2264}-3^{104}$ 
Show that $3^{2264}-3^{104}$ is divisible by $2^6$. 

My attempt: Let $n=2263$. Since $a^{\phi(n)}\equiv 1 \pmod n$ and
$$\phi(n)=(31-1)(73-1)=2264 -104$$
we conclude that $3^{2264}-3^{104}$ is divisible by $2263$. 
I have no idea how to show divisibility by $2^6$.

Note that
$$3^{2264}-3^{104}=3^{104}\cdot (3^{2160}-1)$$
and $\gcd(2,3)=1$, so $64$ divides this number if and only if it divides the factor $3^{2160}-1$. Since 
$$3^{2160}-1=(3^{16})^{135}-1$$
and $3^{16}\equiv 1\bmod 64$, we have that $(3^{16})^{135}\equiv 1^{135}\equiv 1\bmod 64$ and therefore $64\mid (3^{16})^{135}-1$.
