Number theory double powers

Question

$(c.)$ Let $p$ be a prime and $d$ a natural number with $d\mid p-1$ , let $k\in\mathbb{N} , x\in \mathbb{Z}$ and suppose that

$$x^d \equiv 1\mod p^k.$$

Prove there exists a unique $s \in \{\ 0,1,\cdots , p-1\}\ $ such that

$$(x+p^ks)^d \equiv 1 \mod p^{k+1}$$

I can see that X+sP^k is congruent XmoduloP^k but can't arrange to get the power of (k+1) in the modulo term

Answer 1

$(x+p^k s)^d \equiv 1 \pmod{p^k}$.
So $ (x+p^k s)^d \equiv 1 + a_s p^k \pmod{p^{k+1}}$.
Now prove that exactly one $a_s = 0$.

Hint: Which term(s) $\equiv 0 \pmod{p^{k+1}}$?

Since $2k \geq k+1$, hence at most the first 2 terms of the binomial expansion are non-zero. Thus $ (x+p^k s)^d \equiv 1 + a_s p^k \equiv x^d + dx^{d-1} p^ks \pmod{p^{k+1}}$.
Now show that $a_s$ are distinct $\pmod{p}$, so exactly one of them must be $0 \pmod{p}$.
This can be done by contradiction.