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 $2^6$.

You've got $(3^{2160} - 1)\cdot3^{104}$.
The second half of that is obviously not divisible by $2$, so we must show that $3^{2160} \equiv 1\ (\text{mod}\ 2^6)$.
A quick brute force check of exponents of $3$ gives $3^{16} \equiv 1\ (\text{mod}\ 64)$; $2160=16 \times 135$.
