A question about non-archimedean binomial expansion Suppose we work in the number field $\mathbb Q(\omega_{pq})$ with $\omega_{pq}$ a $p q$th root of unity, where $p$, $q$ are distinct primes. Similarly $\omega_{p}$ and $\omega_{q}$ are $p$th and $q$th roots of unit respectively. Now we note the binomial expansion
$$
(1-(1-\omega_{p}))^{\frac{p}{q}}=1-\frac{p}{q}(1-\omega_{p})+ \binom{p/q}2(1-\omega_{p})^{2} + \dotsb.
$$
By simple analysis, we know it should converge in the non-archimedean topology to one of the $q$th roots of unity $\omega_{q}^{k}$ at a certain place $\beta$ over $1-\omega_{p}$, or equivalently over $p$. Denote the limit by $s$. Again by simple analysis of the expansion of $s-1$, we know $s\neq1$.
So we know $1, s, s^{2}, \dotsc, s^{q-1}$ will go once through all the $q$th roots of unity while all of them belong to $1+\pi\mathbb Z_{\beta}$ where $\pi$ is a primal element of place $\beta$. So $1+s+s^{2}+\dotsb+s^{q-1}\in q+\pi\mathbb Z_{\beta}$, so $\neq0$, which is a contradiction of the property of a $q$th root of unity.
What is wrong in the above argument?
 A: What do you mean in the comment to your post that you “get confirmation of $s=1$ from other people”? In any case, it is “simple" to show $s=1$ if you are familiar enough with binomial expansions in $p$-adic fields.
In a complete extension field $K$ of $\mathbf Q_p$, for each element $a$ of $K$ such that $|a-1|_p < 1$, the sequence $\{a^n\}$ for integers $n \geq 0$ is $p$-adically uniformly continuous in $n$ and extends to a $p$-adically (uniformly) continuous function $a^x$ for $x \in \mathbf Z_p$ and $|a^x - 1|_p \leq |x|_p|a-1|_p < 1$. The binomial expansion
$$
a^x = \sum_{k \geq 0} \binom{x}{k}(a-1)^k = 1 + x(a-1) + \binom{x}{2}(a-1)^2  + \cdots 
$$
in $K$ holds for all $x$ in $\mathbf Z_p$.
We have $(a^x)^y = a^{xy}$ for all $x$ and $y$ in $\mathbf Z_p$
since it's true for nonnegative integers $x$ and $y$ and both sides are $p$-adically continuous in $x$ and $y$.  (We have
$|a^x - 1|_p \leq |a-1|_p < 1$ from the binomial expansion
of $a^x$, so $(a^x)^y$ makes sense.)
We'll use this in the complete field $K = \mathbf Q_p(\omega_{pq})$, where $\omega_{pq}$ is a root of unity of order $pq$. Let $a = \omega_p$, so
$|a - 1|_p = |\omega_p - 1|_p = (1/p)^{1/(p-1)} < 1$. (All nontrivial $p$th roots of unity have the same $p$-adic distance from $1$.) For $x$ in $\mathbf Z_p$, what does $a^x = \omega_p^x$ mean? Since $\omega_p^n$ for nonnegative integers $n$ is periodic in $n$ with period $p$, $\omega_p^x$ is a locally constant function: $\omega_p^x = \omega_p^r$ where $x \equiv r \bmod p\mathbf Z_p$ and $0 \leq r \leq p-1$. The binomial expansion above takes the form
$$
\omega_p^x = \sum_{k \geq 0} \binom{x}{k}(\omega_p-1)^k = 1 + x(\omega_p-1) + \binom{x}{2}(\omega_p-1)^2  + \cdots 
$$
for all $x$ in $\mathbf Z_p$. For some reason known only to you, your binomial expansion has $\omega_p - 1$ written in the strange way $-(1-\omega_p)$.  In any case, $p/q$ is a $p$-adic integer
and $p/q \equiv 0 \bmod p\mathbf Z_p$,
so $\omega_p^{p/q} = \omega_p^0 = 1$ in $\mathbf Q_p(\omega_{pq})$ In a similar way, for every nontrivial $p$th root of unity $\omega_p^k$ we have $(\omega_p^k)^{p/q} = \omega_p^{kp/q} = \omega_p^0 = 1$ in $\mathbf Q_p(\omega_{pq})$.
Another strange notation you use is $\beta$ for a place over $p$. I suspect that might be due to thinking the fraktur letter $\mathfrak P$ for a prime ideal is a letter $B$, but it corresponds to $P$. (The fraktur $B$ is $\mathfrak B$.) If you thought $\mathfrak P$ corresponds to $B$, please take the time to rewire your brain to fix that misunderstanding.
