Can we determine $A= 1!+2!+3!+...$'s digits starting from last? After reading a bit about p-adic numbers, I came up with an idea. We know that for every natural number $k$, there exists a natural number $n$ so that for every $m>n$, there are at least $k$ zero digits in the end of $m!$. So we can actually determine $k$ last digits of $1!+2!+3!+...n!$. For example $1!+2!+3!+4!=33$ and $5!=120$ which has $1$ zero at the end. So the last digit becomes fixed with the number $3$ for $n>4$. Similarly, $1!+2!+3!+4!+5!+6!+7!+8!+9!=404073$ and $10!=3628800$. Thus, for $n>9$ the last two digits would be fixed at $73$. So my question is: If we continue this pattern, will we be able to find as many last digits of $A=1!+2!+3!+...$ as we want? Will this be enough to say $A$ is a $10-\text{adic} $ number? To avoid misunderstanding, i would define the last $n$ digits of the number as the first $n$ digits, counting from the right. For example the last digit of $...373737$ is $7$.
 A: Okay let's try to make rigorous sense of this question:
First you need to define what you mean by a 10-adic number (10 is not prime!). The most sensible thing to consider would be, in this context, $R = \varprojlim \mathbb{Z}/10^n\mathbb{Z}$ with the obvious maps. $R$ is profinite and has a natural topology; it is even a topological ring. An open neighborhood basis of $0$ is given by the images of the ideals $(10^k)$ in $R$.
Now you have an infinite series $\sum_{k \geq 0} k!$ and you want to prove that it converges in $R$. Let $S_N = \sum_{k=0}^N k!$ be the partial sum. Define $z_n = [S_{10^n}] \in \mathbb{Z}/10^n\mathbb{Z}$.
As you noticed the $z_n$ are compatible in the sense that $z_{n+1}$ gets mapped to $z_n$ when you project: everything you add after that is a multiple of $10^n$. Therefore they define an element $z \in R$. Then for every $k$, there exists $n_0 = 10^k$ such that $\forall n \geq n_0$, $S_n - z$ is in the neighborhood $(10^k)$ of $0$ in $R$. So $S_N$ converges to $z$.
