Finding remainder of dividing $3^{2^n}$ by $2^{n+3}$ Find the remainder of dividing $3^{2^n}$ by $2^{n+3}$
I was trying to use $ord_{2^{n+3}}(3)$, but i don't see a future in looking for a constant when I change the value of n. However, I was looking for a less advanced solution, is it possible?

You could show by induction that the remainder is $2^{n+2}+1$.
Base case ($n=1$):  $\;3^{2^n}=9=2^{n+2}+1\pmod {2^4=16}$.
Now assume $3^{2^n}\equiv 2^{n+2}+1\bmod 2^{n+3}$; i.e., $3^{2^n}=2^{n+2}+1+k\cdot2^{n+3}$ for some integer $k$.
Then $3^{2^{n+1}}=\left(3^{2^n}\right)^2=2^{2n+4}+1+k^2\cdot2^{2n+6}+2^{n+3}+k\cdot2^{2n+6}+k\cdot2^{n+4}$
$\equiv1+2^{(n+1)+2}\pmod{2^{(n+1)+3}}.$
