For $p\neq 2$ it's easy to prove through the log/exp-correspondence that
$$(1+p\mathbb{Z}_p)^{p^k}=1+p^{k+1}\mathbb{Z}_p.$$
This gives an easy way to compute the groups $\mathbb{Q}_p^*/\mathbb{Q}_p^{*n}$ which are useful in a lot of computations involving local class field theory. I was wondering if there's some nice way to express the above equation in the case $p=2$?
Your equation is equivalent to the following statement: an element $x\in{\mathbb Z}_p$ is congruent to $1\pmod{p^{k+1}}$ if and only if one can write $x=y^{p^k}$, where $y\in{\mathbb Z}_p$ is congruent to $1\pmod p$. I believe you can prove this via Hensel's lemma, starting by looking $\pmod{p^{k+1}}$.
The analogous statement for $p=2$ can, I believe, be proved similarly; it is probably the same statement except with $k+1$ on the right-hand side replaced by $k+2$.
This is related to the fact that the multiplicative groups $({\mathbb Z}/p^k{\mathbb Z})^\times$ are cyclic when $p$ is odd, but $({\mathbb Z}/2^k{\mathbb Z})^\times$ isn't quite: it's the direct product of $\{\pm 1 \bmod 2^k\}$ with a cyclic group. This can be found in any number theory textbook that discusses primitive roots.