For $p\ge 3$ if $g$ is a generator of $\Bbb{Z}/(p^2)^\times$ then it is a generator of all $\Bbb{Z}/(p^k)^\times$. For $a\in \Bbb{Z}_p^\times$ there is a unique $l_{g,k}(a)\in \Bbb{Z}/(p-1)p^{k-1}$ such that $a= g^{l_{g,k}(a)}\bmod p^k$.

For $m<k$, $l_{g,k}(a)= l_{g,m}(a) \bmod (p-1)p^{m-1}$ thus $l_{g,k}(a)= l_{g,m}(a) \bmod p^{m-1}$ and hence $l_g(a)=\lim_{k\to \infty}l_{g,k}(a)$ converges in $\Bbb{Z}_p$ and $a=\lim_{k\to \infty} g^{l_{g,k}(a)}$.

Let $c_k\in 0\ldots p^{k-1}-1,c_k = l_{g,k}(a)\bmod p^{k-1}$. Then $a=\lim_{k\to \infty} g^{c_k}$ iff $a=1\bmod p$. In other words the full discrete logarithm of a $p$-adic number is an element of $\Bbb{Z}/(p-1) \times \Bbb{Z}_p$ and when keeping only the $\Bbb{Z}_p$ part we loose the $a\bmod p$ information.

If $c\in 1+p\Bbb{Z}_p$ then $l_g(c) = \frac{\log_p(c)}{\log_p(g^{p-1})}\in \Bbb{Z}_p$ where $\log_p$ is the $p$-adic logarithm $$\log_p(1+pb)=\sum_{n\ge 1}\frac{p^n(-1)^{n-1} b^n}{n}, b\in \Bbb{Z}_p$$

For $a,a'\in \Bbb{Z}_p^\times$, $l_g(a)=l_g(a')\in \Bbb{Z}_p$ iff $a/a'$ has finite order in $\Bbb{Z}_p^\times$ iff $a^{p-1}=(a')^{p-1}$.

$\log_p(g^{p-1})\ne 1$ when $g$ is an integer. $\log_p(1+pb)=1$ iff $1+pb=\sum_{n\ge 0}\frac{p^n b^n}{n!}=\exp_p(1)$ which is not in $\Bbb{Q}\cap \Bbb{Z}_p$.

$\log_p$ is the discrete logarithm in base $\exp_p(1)$, ie. $\log_p(1+pb)= \lim_{k\to \infty} l_{\exp_p(1)\bmod p^k,k}(1+pb)$.

For $p\ge 3$ if $g$ is a generator of $(\Bbb{Z}/(p^2))^\times$ then it is a generator of all $(\Bbb{Z}/(p^k))^\times$. For $a\in \Bbb{Z}_p^\times$ there is a unique $l_{g,k}(a)\in \Bbb{Z}/(p-1)p^{k-1}\Bbb{Z}$ such that $a= g^{l_{g,k}(a)}\bmod p^k$.

For $m<k$, $l_{g,k}(a)= l_{g,m}(a) \bmod (p-1)p^{m-1}$ thus $l_{g,k}(a)= l_{g,m}(a) \bmod p^{m-1}$ and hence $l_g(a)=\lim_{k\to \infty}l_{g,k}(a)$ converges in $\Bbb{Z}_p$ and $a=\lim_{k\to \infty} g^{l_{g,k}(a)}$.

Let $c_k\in 0\ldots p^{k-1}-1,c_k = l_{g,k}(a)\bmod p^{k-1}$. Then $a=\lim_{k\to \infty} g^{c_k}$ iff $a=1\bmod p$. In other words the full discrete logarithm of a $p$-adic number is an element of $\Bbb{Z}/(p-1) \times \Bbb{Z}_p$ and when keeping only the $\Bbb{Z}_p$ part we lose the $a\bmod p$ information.

If $c\in 1+p\Bbb{Z}_p$ then $l_g(c) = \frac{\log_p(c)}{\log_p(g^{p-1})}\in \Bbb{Z}_p$ where $\log_p$ is the $p$-adic logarithm $$\log_p(1+pb)=\sum_{n\ge 1}\frac{p^n(-1)^{n-1} b^n}{n}, b\in \Bbb{Z}_p$$

For $a,a'\in \Bbb{Z}_p^\times$, $l_g(a)=l_g(a')\in \Bbb{Z}_p$ iff $a/a'$ has finite order in $\Bbb{Z}_p^\times$ iff $a^{p-1}=(a')^{p-1}$.

$\log_p(g^{p-1})\ne 1$ when $g$ is an integer. $\log_p(1+pb)=1$ iff $1+pb=\sum_{n\ge 0}\frac{p^n b^n}{n!}=\exp_p(1)$ which is not in $\Bbb{Q}\cap \Bbb{Z}_p$.

$\log_p$ is the discrete logarithm in base $\exp_p(1)$, ie. $\log_p(1+pb)= \lim_{k\to \infty} l_{\exp_p(1)\bmod p^k,k}(1+pb)$.

For $p\ge 3$ if $g$ is a generator of $\Bbb{Z}/(p^2)^\times$ then it is a generator of all $\Bbb{Z}/(p^k)^\times$. For $a\in \Bbb{Z}_p^\times$ there is a unique $l_{g,k}(a)\in \Bbb{Z}/(p-1)p^{k-1}$ such that $a= g^{l_{g,k}(a)}\bmod p^k$.

For $m<k$, $l_{g,k}(a)= l_{g,m}(a) \bmod (p-1)p^{m-1}$ thus $l_{g,k}(a)= l_{g,m}(a) \bmod p^{m-1}$ and hence $l_g(a)=\lim_{k\to \infty}l_{g,k}(a)$ converges in $\Bbb{Z}_p$ and $a=\lim_{k\to \infty} g^{l_{g,k}(a)}$.

Let $c_k\in 0\ldots p^{k-1}-1,c_k = l_{g,k}(a)\bmod p^{k-1}$. Then $a=\lim_{k\to \infty} g^{c_k}$ iff $a=1\bmod p$. In other words the full discrete logarithm of a $p$-adic number is an element of $\Bbb{Z}/(p-1) \times \Bbb{Z}_p$ and when keeping only the $\Bbb{Z}_p$ part we loose the $a\bmod p$ information.

If $c\in 1+p\Bbb{Z}_p$ then $l_g(c) = \frac{\log_p(c)}{\log_p(g^{p-1})}\in \Bbb{Z}_p$ where $\log_p$ is the $p$-adic logarithm $$\log_p(1+pb)=\sum_{n\ge 1}\frac{p^n b^n}{n}, b\in \Bbb{Z}_p$$

For $a,a'\in \Bbb{Z}_p^\times$, $l_g(a)=l_g(a')\in \Bbb{Z}_p$ iff $a/a'$ has finite order in $\Bbb{Z}_p^\times$ iff $a^{p-1}=(a')^{p-1}$.

$\log_p(g^{p-1})\ne 1$ when $g$ is an integer. $\log_p(1+pb)=1$ iff $1+pb=\sum_{n\ge 0}\frac{p^n b^n}{n!}=\exp_p(1)$ which is not in $\Bbb{Q}\cap \Bbb{Z}_p$.

$\log_p$ is the discrete logarithm in base $\exp_p(1)$, ie. $\log_p(1+pb)= \lim_{k\to \infty} l_{\exp_p(1)\bmod p^k,k}(1+pb)$.

For $p\ge 3$ if $g$ is a generator of $\Bbb{Z}/(p^2)^\times$ then it is a generator of all $\Bbb{Z}/(p^k)^\times$. For $a\in \Bbb{Z}_p^\times$ there is a unique $l_{g,k}(a)\in \Bbb{Z}/(p-1)p^{k-1}$ such that $a= g^{l_{g,k}(a)}\bmod p^k$.

For $m<k$, $l_{g,k}(a)= l_{g,m}(a) \bmod (p-1)p^{m-1}$ thus $l_{g,k}(a)= l_{g,m}(a) \bmod p^{m-1}$ and hence $l_g(a)=\lim_{k\to \infty}l_{g,k}(a)$ converges in $\Bbb{Z}_p$ and $a=\lim_{k\to \infty} g^{l_{g,k}(a)}$.

Let $c_k\in 0\ldots p^{k-1}-1,c_k = l_{g,k}(a)\bmod p^{k-1}$. Then $a=\lim_{k\to \infty} g^{c_k}$ iff $a=1\bmod p$. In other words the full discrete logarithm of a $p$-adic number is an element of $\Bbb{Z}/(p-1) \times \Bbb{Z}_p$ and when keeping only the $\Bbb{Z}_p$ part we loose the $a\bmod p$ information.

If $c\in 1+p\Bbb{Z}_p$ then $l_g(c) = \frac{\log_p(c)}{\log_p(g^{p-1})}\in \Bbb{Z}_p$ where $\log_p$ is the $p$-adic logarithm $$\log_p(1+pb)=\sum_{n\ge 1}\frac{p^n(-1)^{n-1} b^n}{n}, b\in \Bbb{Z}_p$$

For $a,a'\in \Bbb{Z}_p^\times$, $l_g(a)=l_g(a')\in \Bbb{Z}_p$ iff $a/a'$ has finite order in $\Bbb{Z}_p^\times$ iff $a^{p-1}=(a')^{p-1}$.

$\log_p(g^{p-1})\ne 1$ when $g$ is an integer. $\log_p(1+pb)=1$ iff $1+pb=\sum_{n\ge 0}\frac{p^n b^n}{n!}=\exp_p(1)$ which is not in $\Bbb{Q}\cap \Bbb{Z}_p$.

$\log_p$ is the discrete logarithm in base $\exp_p(1)$, ie. $\log_p(1+pb)= \lim_{k\to \infty} l_{\exp_p(1)\bmod p^k,k}(1+pb)$.