prove or disprove: $((-2)^{(p-1)/2})^{p^{k-1}}\equiv \pm 1\bmod p^k\Leftrightarrow (-2)^{(p-1)/2} \equiv \pm 1 \bmod p$ 
Prove or disprove: if $p$ is an odd prime number and $k\ge 1$, then $((-2)^{(p-1)/2})^{p^{k-1}}\equiv 1\bmod p^k\Leftrightarrow (-2)^{(p-1)/2} \equiv 1\bmod p$  and $((-2)^{(p-1)/2})^{p^{k-1}}\equiv -1\bmod p^k\Leftrightarrow (-2)^{(p-1)/2} \equiv -1\bmod p$.

The reverse implication in both cases seems to hold, but I'm not sure how to prove the forward implication in either case. For the reverse implication for the first case, if $(-2)^{(p-1)/2}\equiv \pm 1\bmod p,$ then $(-2)^{(p-1)/2})^{p^{k-1}} = (xp \pm 1)^{p^{k-1}}$ for some $x\in \mathbb{Z}$, and by the Binomial theorem it is easy to see that this expression is equivalent to $\pm 1$ modulo $p^k$.
 A: We assume the following:
$$((-2)^{(p-1)/2})^{p^{k-1}} \equiv \pm 1 \pmod {p^k}$$
It now rests us to show that (where $\pm$ should be chosen the same in the two equations):
$$(-2)^{(p-1)/2} \equiv \pm 1 \pmod p$$
We will take the first equation modulo $p$ (we may since $p^k \equiv 0 \pmod p$) and get:
$$((-2)^{(p-1)/2})^{p^{k-1}} \equiv \pm 1 \pmod {p}$$
Since $p$ is an odd prime number we know that:
$$(-2)^{(p-1)/2} \in \{ -1, +1 \} \pmod p$$
Also, since $p$ is odd we get that $p^{k-1}$ is odd, thus:
$$((-2)^{(p-1)/2})^{p^{k-1}} \equiv + 1 \pmod {p} \implies (-2)^{(p-1)/2} \equiv +1 \pmod{p}$$
and
$$((-2)^{(p-1)/2})^{p^{k-1}} \equiv - 1 \pmod {p} \implies (-2)^{(p-1)/2} \equiv -1 \pmod{p}$$
Exactly what we wanted to show.
A: We suppose $k\ge 1$.
$$((-2)^{(p-1)/2})^{p^{k-1}} \equiv \pm 1 \pmod {p^k} \implies ((-2)^{(p-1)/2})^{p^{k-1}} \equiv \pm 1 \pmod {p}.$$
By Fermat little Theorem:
$$((-2)^{(p-1)/2})^{p^{k-1}} \equiv  ((-2)^{(p-1)/2})^{p^{k-2}}\equiv\cdot\cdot\cdot  \equiv(-2)^{(p-1)/2}\pmod {p}.$$
Then $$((-2)^{(p-1)/2})^{p^{k-1}} \equiv \pm 1 \pmod {p^k} \implies (-2)^{(p-1)/2} \equiv \pm 1 \pmod {p}.$$
