A theorem on p-adic power series This is a question about a proof of a theorem in many books on $p$-adic numbers and I don't seem to understand one of the directions. The theorem is
Let $f(X) = \sum\limits_{n=0}^\infty a_nX^n \in \mathbb{Q}_p[[X]]$ be a $p$-adic power series. Let $c \in D_f$, the domain of convergence for $f$, so $f(c)$ converges. Define $g_m$ by
$g_m = \sum\limits_{n \geq m} \left(\begin{array}{c} n\\m\end{array} \right)a_nc^{n-m}$
and set $g(X) = \sum\limits_{m=0}^\infty g_mX^m.$ Then $D_f = D_g$ (the domain of convergence of $g$) and further, for all $b \in D_f$, we have $f(b+c) = g(b).$
I have no problem with the proof that $D_f \subseteq D_g$, it's the converse I don't understand. Every resource I've gone to has said the argument is symmetric and follows from reversing the roles of $f$ and $g$. I don't see why this is true. The whole argument to show that $D_f \subseteq D_g$ is given by controlling the term 
$\left| \left(\begin{array}{c} n\\m\end{array} \right)a_nc^{n-m}b^m\right|_p$
where $b \in D_f$. But when I try to do this for $b \in D_g$, I can't get the same bound. In fact, what I get is that the above term is bounded by $|a_n|_p\rho^n$ where $\rho$ satisfies $|c|_p \leq \rho$ and $|b|_p \leq \rho$ and I see no reason why this term should go to zero, and further I don't see how this plays off of any symmetry in the proof. Any help to get my understanding what I'm misinterpreting would be greatly appreciated.
 A: That must surely be the least intuitive way possible of stating that simple result. To understand what’s happening, you should do some hand computation on series. Let me show you:
If you expand out $f(X+c)$, just go as far as, maybe the $X^4$-terms, you’ll see that $g(X)=f(X+c)$. Now, the domain of convergence of $f$ is a group $S$, and your hypothesis is that $c$ is in $S$. Since $S-c=S$, the numbers that make $g$ converge are exactly the numbers that make $f$ converge. That’s all there is to it. Of course you do have to justify my statement “you’ll see that...”.
EDIT — Expansion:
If you believe your own proof that $D_f\subseteq D_g$, and that for every $z\in D_f$, $f(z+c)=g(z)$, then it still seems to me that you’re done. Let $f(X+c)=\sum_{n\ge0}b_nX^n=g(X)$, and do the same thing in reverse, $h(X)=g(X-c)=\sum_{n\ge0}\beta_nX^n$, so by your result, $D_g\subseteq D_h$. Now from the evaluation part, $h(X)-f(X)$ is a power series that vanishes at every $z\in D_f$, and thus is the zero power series, so that $h=f$, $D_h=D_f$, and thus $D_g=D_f$.
