How to prove that $a^{2^{n-2}} \equiv 1 \pmod{2^n}$? 
Let $a$ be an odd integer and $n$ an integer such that $n\ge 3$. 
1) I want to show that $a^{2^{n-2}} \equiv 1 \pmod{2^n}$
2) Then I want to show that $(\mathbb Z/{2^n\mathbb Z})^*$, the multiplicative group of integers modulo $2^n$,  is not cyclic.

My Try: 
1) Since $\phi(2^{n-1})=2^{n-2}$ and that $a$ is coprime with $2^{n-1}$, then by Euler's theorem we have that
$a^{2^{n-2}} \equiv 1 \pmod{2^{n-1}}$ but I can't write the final conclusion.
2) I know that we have to show that no single element can generate the whole group but I don't know how to do it.
Thank you for your help!!
 A: Hints:
Suppose that you've a congruence like $f(x) \equiv a \pmod{p^m}$.
Do you know how you can construct a solution from this congruence for the congruence $f(x) \equiv a \pmod{p^{m+1}}$ ? 
You have to use the Taylor formula for the expansion of $f(x+y)$ where $f$ is a polynomial. See what you get.
EDIT: Actually in this particular case, you can do it more easily by writing $a^{2^{k-2}} = 1 + t2^k$ and then applying the binomial theorem... But it's essentially the same thing and it's generally good to know how you can find the solutions of a congruence modulo a prime power provided that you've solved it for the previous exponent of that prime number.
For the second part, if $g$ is a generator of the group $(\mathbb{Z}/n\mathbb{Z})^\times$ where $n=2^k, k>2$, then the order of $g$ must be equal to the order of the group. So...
In elementary number theory, the fact that $(\mathbb{Z}/n\mathbb{Z})^\times$ is not cylic for $n=2^k, k>2$ is stated as there exist no primitive roots modulo $n$ for exponents of $2$ higher than three.
A: Hints:
For 1), start with $a^2 \equiv 1 \pmod{8}$ and induct on $n \geq 3$.
For 2), take a closer look at what you are proving in 1). What is the order of the group? You have already shown what you wanted to show: that no element generates the whole group.
