On page 15 of these notes by Colmez, he claims that for $a \in \mathbb{Z}^*_p$ (i.e. $a$ is a unit in $\mathbb{Z}_p$), the following power series is in $\mathbb{Z}_p[[T]]$ and has "valuation" equal to $0$:
$$ \frac{1}{T}-\frac{a}{(1+T)^a-1} $$ where the valuation $\nu$ of a power series $\sum_{n=0}^{\infty}a_nT^n \in \mathbb{Z}_p[[T]]$ is defined as $$\nu \left(\sum_{n=0}^{\infty}a_nT^n \right)=\inf_{n}\nu_p(a_n)$$ $\nu_p$ being the usual $p$-adic valuation on $\mathbb{Z}_p$: $\nu_p(x) = \sup \{n\in\mathbb{Z} \mid x\in p^n\mathbb{Z}_p\}$. I have showed that the power series is in $\mathbb{Z}_p[[T]]$, but can't show that the valuation is $0$. I tried manually calculating the coefficients, but that doesn't lead to anywhere. Any help would be appreciated.
The power series we're looking at is, per a standard geometric series argument,
$$T^{-1} \cdot (x-x^2+x^3-x^4+x^5- \dots)$$
where $x$ is itself the series
$$x= a^{-1}\sum_{n=2}^\infty \binom{a}{n} T^{n-1} = \frac{a-1}{2}T + \frac{(a-1)(a-2)}{6} T^2 + \dots$$
It suffices to show that one of the coefficients of $T$-powers in the series $x$ has valuation $0$. (Either by believing that the valuation $\nu$ of power series is a well-defined ultrametric valuation, or checking by hand: If there is a smallest power of $T$ in $x$ whose coefficient has valuation zero, then by the ultrametric principle the valuations of the corresponding $T$-power in $x^2, x^3, ...$ are all greater than zero, hence by the same principle this $T$-power has coefficient of valuation $0$ in $x-x^2+x^3 - \dots$.)
But that is straightforward for $a \in \mathbb Z_p^\times \setminus\{1\}$:
If $v_p(a-1)=0$ (note this implies $p \neq 2$), then already the first one, $a^{-1} \binom{a}{2}$, does the job.
If $v_p(a-1)=r \in \{1,2,3,\dots\}$, then check that $v_p(\binom{a}{p^r}) =0$.
As said, in the case $a=1$ (i.e. $r=\infty$) the statement is false, the series in question becomes $0$.