How to prove the p-adic expansion of negative p-adic integers ends with p-1?

Question

I just started calculating with p-adic numbers and one of the questions I'm stuck on is how I can prove that for a negative integer the p-adic expansion ends with an infinite amount of digits p-1. I tried to check it with some examples like $-1 = ...(p-1)(p-1)(p-1)$ and $-2 = ... (p-1)(p-1)2$ and so on. I tried proving it with induction on $n$, but I don't see how I can complete the induction step. I thougt about $-n = -(n-1)-1$, but I'm confused with how I can add these up to see $-n$ again ends with $p-1$.

Answer 1

See Case 1 in the proof of Theorem 3.1 here.

For a self-contained description, first look at what an infinite $p$-adic expansion is when all of its digits are $p-1$ and it starts in position $p^d$: $$ \sum_{k \geq d} (p-1)p^k = (p-1)\frac{p^d}{1-p} = -p^d. $$

With this in mind, when $n$ is a negative integer, pick $d$ so large that $-p^d < n < 0$. Then $N := p^d + n$ has $0 < N < p^d$, so when we write $N$ in base $p$ its digits will involve nothing higher than $p^{d-1}$: $$ N = c_0 + c_1p + \cdots + c_{d-1}p^{d-1} $$ where $0\leq c_i \leq p-1$. Then $$ n = N - p^d = c_0 + c_1p + \cdots + c_{d-1}p^{d-1} - p^d = c_0 + c_1p + \cdots + c_{d-1}p^{d-1} + \sum_{k \geq d} (p-1)p^k. $$

Answer 2

(Edit: Replaced earlier answer with different approach.)

Your expansion of $-2$ is incorrect. I would think if you see the expansions of negative numbers, you'll see what is happening. Let's take $p=5$ for example, then

$$-1 = ...44444444$$ $$-2 = ...44444443$$ $$-3 = ...44444442$$ $$-4 = ...44444441$$ $$-5 = ...44444440$$ $$-6 = ...44444434$$ $$-7 = ...44444433$$ $$-8 = ...44444432$$ $$-9 = ...44444431$$ $$-10 = ...44444430$$ $$-11 = ...44444424$$ $$\dots$$ $$-24 = ...44444401$$ $$-25 = ...44444400$$ $$-26 = ...44444344$$ $$\dots$$ See how you will not run out of $4$'s on the left.