Let $p$ be a prime.
It is basic that following isomorphisms;
$\mathbb{Z}_p^{\times}/(1+p^n\mathbb{Z}_p )\cong (\mathbb{Z}_p/p^n\mathbb{Z}_p)^{\times} $,
$(1+p^n\mathbb{Z}_p)/(1+p^{n+1}\mathbb{Z}_p )\cong \mathbb{Z}_p/p\mathbb{Z}_p$
But I don’t figure out why those split i.e.
(1) $\mathbb{Z}_p^{\times}\cong(1+p^n\mathbb{Z}_p )\times(\mathbb{Z}_p/p^n\mathbb{Z}_p)^{\times} $,
(2) $1+p^n\mathbb{Z}_p\cong(1+p^{n+1}\mathbb{Z}_p )\times \mathbb{Z}_p/p\mathbb{Z}_p$.
To get this off the unanswered list: Except in very few exceptional cases, neither of your isomorphisms (1) nor (2) is true. I.e. those sequences, in general, do not split.
The few cases where they do split are exhibited by Johann Birnick in a comment.
In fact, the only torsion elements i.e. roots of unity in the multiplicative group $\mathbb Q_p^\times$ are well-known to be
$\{\pm 1\}$ for $p=2$, and $\mu_{p-1}$ (the $p-1$-th roots of unity) for odd $p$.
But the group on the RHS of you proposed iso (1) has elements of order $p^n-p^{n-1}$, and the group on the RHS of your proposed iso (2) has elements of order $p$.
This excludes iso (1) for $n\ge 2$ if $p$ odd, for $n \ge 3$ if $p=2$; and it excludes iso (2) for all odd $p$.
Finally, to exclude iso (2) for $p=2$ and $n \ge 2$, just note that the torsion element $-1$ is not in $1+2^n \mathbb Z_2$ for any $n \ge 2$, i.e. that subgroup cannot contain more torsion elements than the trivial $+1$, certainly cannot have a subgroup isomorphic to $\mathbb Z/2$.