Show that $|5|=2^{n-2}$ when $n \ge 3$ in $U(2^n)$.
I wrote $$5^{2^{n-2}}=\left(1+2^2\right)^{2^{n-2}}=1+2^{2}.2^{n-2}+\frac{2^{n-2}.(2^{n-2}-1)}{2!}2^4+....+\frac{2^{n-2}.(2^{n-2}-1).(2^{n-2}-2)..(2^{n-2}-k+1)}{k!}2^{2k}+..+{2^{2}}^{2^{n-2}}$$
In mod$(2^n)$, all the terms should go to $0$ except $1$ which is the first term.
It does hold for $n=3$ and $n=4$ .... I am not able to prove it in general case
Thanks for the help!!
One way to deal with this is to use the identity $$a^2-b^2=(a+b)(a-b).$$ So that $$5^{2^{n-2}}-1=(5^{2^{n-3}}+1)(5^{2^{n-3}}-1)=\cdots.$$ Can you continue from here?