Looking for all $p$ power cyclotomic units of $Q_p$ or the generator? This is related to decomposition of $U_1=1+pZ_p$ in Iwasawa's Local class field theory book Chpt 2, Proposition 2.7
Let $k$ be a finite $Q_p$ extension, $d=[k:Q_p]$ and $q$ cardinality of residue field of $k$. Set $\pi$ uniformizer of valuation ring associated to $k$. Then $k^\times=<\pi>\oplus Z_{(q-1)}\oplus Z_{p^a}\oplus Z_p^d$ where $Z_{p^a}$ corresponds to all $p$-power roots of unity and $U_1=Z_p^d\oplus Z_{p^a}$. However $a$ is not concretely given. 
$\textbf{Q:}$ I want to determine numerical values of $a$. The proof is not using constructive method by uses PID classification theorem and via $|U_1/U_n|<\infty$ for some large $n$ s.t. $U_n\cap W=1$ where $U_n$ is free of rank $d$ over $Z_p$(p-adic integer). The latter does not indicate exact size of $W$. How do I find this $a$ concretely in terms of $e,f,d$?
 A: $$K^* = O_K^\times \times \pi^\Bbb{Z}$$
$$O_K^\times= \langle \zeta_{p^f-1}\rangle \times 1+\pi O_K$$
$$H = (1+\pi O_K)_{tors}$$
Since $H$ is torsion it contains some roots of unity, since they are equal to $1\bmod (\pi)$ it means they are $p^r$-th roots of unity, ie. $H = \langle \zeta_{p^a}\rangle$.

Because $\lim_{n\to \infty} (1+\pi x)^{p^n}=1$ then $(1+\pi x)^b$ is well-defined for $b\in \Bbb{Z}_p$ and $(1+\pi O_K)/H$ is a torsion free $\Bbb{Z}_p$-module, it is finitely generated because $1+p^2 O_K$ is of finite index in $1+\pi O_K$ and $\log$ is a $\Bbb{Z}_p$-module isomorphism $1+p^2 O_K\to p^2 O_K$, 

that is to say $$(1+\pi O_K)/H=  \prod_{j=1}^{[O_K:\Bbb{Z}_p]} (1+\pi x_j)^{\Bbb{Z}_p}  H$$
$$1+\pi O_K = H \times \prod_{j=1}^{[O_K:\Bbb{Z}_p]} (1+\pi x_j)^{\Bbb{Z}_p} $$
$p^f$ is the size of $O_K/(\pi)$ and $\zeta_{p^a}$ is the largest $p^r$-root of unity in $K$.
A: The non-$p$-primary part of the group of roots of unity of a local $p$-adic field $K$ is completely known thanks to Hensel's lemma: it's  $\mu_{p^f}-1$, where $f$ is the inertia index over $\mathbf Q_p$. As for the $p$-primary part $\mu_{p^a}=<\zeta_{p^a}>$ (with $p^a$ as in your notation), it depends on $K$, as stressed by @Lubin. But although no all-purpose formula can exist, a general upper bound can be given. Denote by $U_n$ the multiplicative group $1+P^n$, where $P$ is the maximal ideal of the ring of integers of $K$. Let $M$ be the valuation of $\zeta_{p^a}-1$. If $e$ is the ramification index of $K/\mathbf Q_p$ and $e'=e/p-1$, then $M\le e'$. This follows from the classical property that the map $x\to x^p$ is an isomorphism of $U_n$ onto $U_{n+e}$ if $n>e'$ (see e.g. Serre's "Local Fields", XIV, prop. 9), hence in particular these groups have the same torsion.
