Expressing $ (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ in terms of powers of $5$ I was doing my preview while some problems turned out to make me feel sucky.
It's said that every element $ x $ of $ (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ could be expressed uniquely as
$$ x \equiv\pm 5^{\alpha} \bmod 2^k $$
where $ 1\le\alpha\le 2^{k-2} $, and the exponent $$ e=2^{k-2} $$
which I understood as the $ l.c.m $ of the orders of all $ x \in (\mathbb{Z}/2^k\mathbb{Z})^{\times} $ equals to $ 2^{k-2} $
All these above were written in my teacher's PPT for the number theory class. Since I'm doing a preview, I don't quite get that. Why is it and how to understand it?
Any explanation would be greatly appreciated!
 A: The following is an excerpt from a book whose draft I wrote last year. It answers your question in essentially the same way as in the comments by reuns, but in explicit detail.
Problem. Take the following steps for all integers $n\ge 3$:

*

*Prove by induction on $n$ that $\nu_2 \left(5^{2^{n-2}}-1\right) =
    n.$

*Show that the $\text{ord}_{2^n}(5) = 2^{n-2}.$

*Deduce that
$S=\left\{\pm 5^k : k\in \left[2^{n-2}\right] \right\}$ is a reduced
residue system modulo $2^n.$
Solution. Each step leads to the next one.

*

*We will show by induction on $n\ge 3$ that, for each integer $n\ge 3,$ there exists an odd integer $x_n$ such that $5^{2^{n-2}} = 1 + x_n \cdot 2^n.$ Since $x_n$ will be shown to be odd, no power of $2$ higher than $2^n$ can divide $5^{2^{n-2}}-1,$ while it is true that $2^n$ does divide $5^{2^{n-2}}-1.$ Thus, we will have proven that $\nu_2 \left(5^{2^{n-2}}\right) = n.$
The base case $n=3$ holds because $$5^{2^{3-2}} - 1 = 24 = 3\cdot 2^3.$$ Now suppose $5^{2^{n-2}} = 1 + x_n \cdot 2^n$ for some integer $n\ge 3$ and odd integer $x_n.$ Squaring the equation yields
$$5^{2^{n-1}} = 1 + x_n \cdot 2^{n+1} + x_n^2 \cdot 2^{2n} = 1 + x_n(1+x_n 2^{n-1})\cdot 2^{n+1}\equiv 1\pmod{2^{n+1}}.$$ Since $x_n$ is odd and $n\ge 3,$ so is $$x_{n+1}=x_n(1+x_n 2^{n-1}).$$


*We know from the first part that $$5^{2^{n-2}} \equiv 1\pmod{2^n}.$$ So the order of $5$ modulo $2^n$ must divide $2^{n-2}.$ Suppose, for contradiction, that there exists an integer $i$ such that $3\le i\le n$ and $5^{2^{n-i}}\equiv 1\pmod{2^n}.$ Squaring this sufficiently many times (taking it to an exponent of $2^{i-3},$ to be precise) we get $5^{2^{n-3}} \equiv 1\pmod{2^n}.$ So there exists an integer $y_n$ such that $$5^{2^{n-3}} = 1 + y_n 2^n.$$ Squaring this yields $$5^{2^{n-2}} = 1 + y_n 2^{n+1} + y_n^2 2^{2n} = 1+(2y_n + y_n^2 2^n)2^n,$$ which implies that the $x_n$ from the last part is the even number $2y_n + y_n^2 2^n$. This is a contradiction.


*Note that the order of $5$ modulo $2^n$ is $2^{n-2}=\frac{\varphi(2^n)}{2}$ and that the elements of $S$ are all in fact units. We could show that $S$ consists of precisely all of the units (that is, invertible elements, which are the odd integers in this case) by showing that none of the $5^i$ coincide with the $-5^j$ modulo $2^n.$ If $5^i\equiv -5^j\pmod {2^n}$ then cancelling the smaller of the two powers from both sides (or either, if they are equal) would yield a power of $5$ congruent to $-1$ modulo $2^n.$ It suffices to show that $$5^k \not\equiv -1\pmod{2^n}$$ for all $k.$ If the congruence held, then we could reduce it to $$5^k \equiv -1 \equiv 3\pmod{4}$$ using $4\mid 2^n.$ This is contradictory because $$5^k \equiv 1^k \equiv 1\pmod{4}.$$ Therefore, there are no overlaps between the $5^i$ and $-5^j.$ This produces $$2\cdot 2^{n-2}=2^{n-1}=\varphi(2^n)$$ distinct units modulo $2^n,$ which is the maximal number and so this is the set of all units modulo $2^n.$
