Systematic way to solve this congruence

Question

If I have prime $p$ and some arbitrary positive integer $k$, is there a closed form for the multiplicative inverse of $p-1$ modulo $p^k$?

Answer 1

First, the inverse clearly exists, since $\gcd(p-1,p^k)=1$.

$$1\equiv -\left(p^k-1\right)\pmod{p^k}$$

Divide both sides by $p-1$:

$$\iff \frac{1}{p-1}\equiv -\left(p^{k-1}+p^{k-2}+\cdots+1\right)\pmod{p^k}$$