Prove that $\mathbb{Z}_{2^k}^* \cong \mathbb{Z}_2 \times \mathbb{Z}_{2^{k-2}}$

Question

In order to compute the value of the Carmichael function for $2^k$ with $k > 2$, my notes indicate that one has to stablish that $\mathbb{Z}_{2^k}^* \cong \mathbb{Z}_2 \times \mathbb{Z}_{2^{k-2}}$ .

However, I have not been able to find this theorem in the bibliography (essentially Shoup's number theory), nor googling nor on stack exchange.

Does anyone know proof of this fact?

Edit

Here is a python script for the powers I was asked to compute:

for i in range(3,40):
    find = True
    for j in range(1,280):
        if(5**j % 2**i == 1 and find == True):
            find = False
            print(2**i," has finished with ", j)

and here the results:

8  has finished with  2
16  has finished with  4
32  has finished with  8
64  has finished with  16
128  has finished with  32
256  has finished with  64
512  has finished with  128
1024  has finished with  256

So experimentally, it seems that $ord(5) = 2^{k-2}$ for $k > 2$.

Answer 1

Hint: compute the order of $5$ in $(\mathbb{Z}/(2^k))^{\times}$.

Edit: I was asked to be more specific. I suggest that you prove that $2^k$ does not divide $5^{2^{k-3}}-1$ but divides $5^{2^{k-2}}-1$. Thus, you can deduce that $(x,y) \in \mathbb{Z}/(2) \times \mathbb{Z}/(2^{k-2}) \rightarrow (-1)^x5^y \in (\mathbb{Z}/(2^k))^{\times}$ is an isomorphism.