An identity of the exponential series: $\text{exp}(x) = 1 + c$ 
Let $K_p$ be the completion of an algebraic number field at a prime ideal $p$, with valuation $v_p$. Define
  $$\text{exp}(x) = \sum_{i=0}^\infty \frac{x^i}{i!} $$ 
  for $x \in K_p$. 

How does one make sense of this definition? If $x \in K_p = \lim K/p^n$ is some coherent sequence, what does it mean exactly to take the infinite sum? Of course, the inverse limit is a ring so I guess the powering is well-defined and $i!$ is the image of under the map $\mathbf{Z}$? 

For $x \in K_p$ satisfying $v_p (x) > \frac{v_p (p)}{p-1} $, we have $\text{exp}(x) = 1+ c$ with $v_p(c) = v_p(x)$

Once again, how do we make sense of the equality? I guess this is an equality in an inverse limit $K_p$. My intuition for explicit computations in such does not seem to be serving me well here though. 
The idea of the proof involves showing that all of the terms of the series after $1 + x$ lie in a higher power of $p$ than does $x$. Why does this show equality?
 A: You should drop the proj. lim. and work instead with convergent series in $\mathbf C_p$, the $p$-adic completion of the algebraic closure of $\mathbf Q_p$. The point is that in non archimedean analysis, a series converges iff it general term tends to zero (an analyst's dream !). Thus:


*

*For the convergence of the series $\exp(x)$, all you actually need is an estimate of $v_p (n!)$. Since there are $[n/p^i]$ multiples of $p^i$ less than $n!$, it is plain that the exponent of $p$ in $n!$ is $[n/p]+[n/p^2]+\dotsb< n/(p-1)$. If $p^a\le n < p^{a+1}$, then the sum is larger than $n/p +\dots+ n/p^a - a = n/(p-1) - a - np^{-a}/(p-1) > (n-p)/(p-1) - \log (n)/\log (p)$, hence $(n-p)/(p-1) - \log (n)/\log (p) < v_p (n!)<n/(p-1)$. It follows that $|x^n/n!| \to 0$ if $|x|<p^{-1/(p-1)}$ and $|x^n/n!|\to \infty$ if $|x|>p^{-1/(p-1)}$. This shows that the radius of convergence of the series $\exp(x)$ is $p^{-1/(p-1)}<1$ .

*For the series $\log (1+x)$ , since $|x^n/n!|<1$, one checks that $|\exp(x)-1|<1$ if $|x|<p^{-1/(p-1)}$, so $\exp(x)$ and $\log(\exp(x))$ converge and verify $\log(\exp(x))=x$ and $\exp(\log(1+x))=1+x$ if $|x|<p^{-1/(p-1)}$.
You can find all the details e.g. in Washington's book, §5.1.    
A: Assume that $K=\mathbb{Q}$ for simplicity, so that $K_p$ is the $p$-adic numbers $\mathbb{Q}_p$. The $p$-adic absolute value is related to the valuation by
$$ |x|_p=p^{-v_p(x)}$$
and is non-Archimedean, i.e. $|x+y|_p\leq \max\{|x|_p,|y|_p\}$ for all $x,y\in\mathbb{Q}_p$. This, together with the Cauchy criterion for series, implies that $\sum_{n=1}^{\infty}a_n$ converges if and only if $\lim_{n\to\infty}a_n=0$.
Now consider $\exp(x)$. Using de Polignac's formula, one can show that
$$ v_p(n!)=\frac{n-S(n)}{p-1}$$
where $S(n)$ is the sum of the digits in the base $p$ expansion of $n$.
Hence if $v_p(x)>\frac{1}{p-1}$ then 
$$ v_p\Big(\frac{x^n}{n!}\Big)=nv_p(x)-v_p(n!)\geq n\Big(v_p(x)-\frac{1}{p-1}\Big)$$
so $\lim_{n\to\infty}\frac{x^n}{n!}=0$ and $\exp(x)$ converges.
For the last part, recall that if $v_p(x)\neq v_p(y)$ then in fact $v_p(x+y)=\min\{v_p(x),v_p(y)\}$. Therefore if $v_p(x)<\frac{1}{p-1}$ then
$$ \exp(x)=1+\sum_{n=1}^{\infty}\frac{x^n}{n!}=1+c$$
and $v_p(c)=v_p(x)$ because if $n\geq 2$ then $S(n)\geq 1$, hence
$$v_p\Big(\frac{x^n}{n!}\Big)=nv_p(x)-\frac{n-S(n)}{p-1}\geq v_p(x)+(n-1)\Big(v_p(x)-\frac{1}{p-1}\Big)>v_p(x) $$
