I want to understand the group structure of the group of units in the ring $\mathbb{Z}/2^m \mathbb{Z}$ for positive integers $m$. I expect this has been answered before here on MSE, but so far the automatic suggestions didn't turn it up.
Here is what I understand: it is an abelian group, hence a product of cyclic ones. Also the order of this group is $2^{m-1}$ so that the orders of the factors in the product are all powers of two by Lagrange's theorem.
Now these conditions are restrictive enough to compute the first few cases by hand.
We have $(\mathbb{Z}/2 \mathbb{Z})^* = C_1$, $(\mathbb{Z}/4 \mathbb{Z})^* = C_2$, $(\mathbb{Z}/8 \mathbb{Z})^* = C_2 \times C_2$, $(\mathbb{Z}/16 \mathbb{Z})^* = C_2 \times C_4$, $(\mathbb{Z}/32 \mathbb{Z})^* = C_2 \times C_8$.
I'm starting to believe that for $m > 1$ we have that $(\mathbb{Z}/2^m \mathbb{Z})^* = C_2 \times C_{2^{m-2}}$ but is there a simple conceptual explanation for that (if true)?