Show that $(x+1)^p \equiv x^p +1 \pmod{p^2}$ given conditions on $x$ and $p$. The problem is as follows.

Let $p$ be a prime of the form $p = 3k + 1$ for some $k \in \mathbb{Z}$. Let $x$ be an integer such that $p \nmid x$ and $[x]_p$ has multiplicative order 3 in the ring $(\mathbb{Z}/p\mathbb{Z})^{\times}$. Show that
$$
(x+1)^p \equiv x^p+1 \pmod{p^2}
$$

My attempt:
Rewrite $(x+1)^p$ in its binomial expansion. We get $(x+1)^p = x^p+ {p\choose 1}x^{p-1} + {p\choose 2}x^{p-2} + \dots + {p\choose p-1}x + 1$. Now it is sufficient to prove that ${p\choose 1}x^{p-1} + {p\choose 2}x^{p-2} + \dots + {p\choose p-1}x \equiv 0 \pmod {p^2}$. However I do not know how to prove this last equivalence.
Any hints on how to procede would be greatly appreciated!
My knowledge consists of basic number theory and $p$-adic numbers and related theorems.
 A: I don't offhand see any way to use the binomial expansion to solve your question. Instead, here's an alternate approach. Because $[x]_p$ has multiplicative order $3$, then $x \not\equiv 1 \pmod{p}$ and
$$x^3 \equiv 1 \pmod{p} \implies x^3 - 1 \equiv 0 \pmod{p} \tag{1}\label{eq1A}$$
Note the lifting-the-exponent lemma states that, for odd primes $p$ where $p \not\mid x$ and $p \not\mid y$, then


*

*If $p \mid x - y$, $\; \nu _{p}(x^{n}-y^{n}) = \nu _{p}(x - y) + \nu _{p}(n)$.

*If $n$ is odd and $p \mid x + y$, $\; \nu _{p}(x^{n} + y^{n}) = \nu _{p}(x + y) + \nu _{p}(n)$.


Thus, using \eqref{eq1A} and the first option above gives
$$\nu_p((x^{3})^{p} - 1) = \nu_p(x^3 - 1) + \nu_p(p) \ge 2 \tag{2}\label{eq2A}$$
We therefore have
$$x^{3p} - 1 \equiv 0 \pmod{p^2} \tag{3}\label{eq3A}$$
Using Fermat's little theorem gives $x^{p} - 1 \equiv x - 1 \not\equiv 0 \pmod{p}$. Since $x^{3p} - 1 = (x^p - 1)(x^{2p} + x^{p} + 1)$, we thus get from \eqref{eq3A} that
$$x^{2p} + x^{p} + 1 \equiv 0 \pmod{p^2} \implies x^{p} + 1 \equiv -x^{2p} \pmod{p^2} \tag{4}\label{eq4A}$$
Since $x^3 = (x - 1)(x^2 + x + 1)$ and $x \not\equiv 1 \pmod{p}$, we thus get from \eqref{eq1A} that
$$x^2 + x + 1 \equiv 0 \pmod{p} \tag{5}\label{eq5A}$$
Next, using the second option of the lifting-the-exponent lemma statement mentioned earlier gives
$$\nu_p((x^{2})^{p} + (x + 1)^p) = \nu_p(x^2 + x + 1) + \nu_p(p) \ge 2 \tag{6}\label{eq6A}$$
From this and using \eqref{eq4A}, we get
$$\begin{equation}\begin{aligned}
x^{2p} + (x + 1)^p & \equiv 0 \pmod{p^2} \\
(x + 1)^p & \equiv -x^{2p} \pmod{p^2} \\
(x + 1)^p & \equiv x^p + 1 \pmod{p^2}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
A: Since $x^3=1\mod p$ it is p-adically near the Teichmuller character $\omega^3=1$ in $\mathbb{Q}_p$ and so we may write it as, $x=\omega+pz$ for some p-adic integer $z$.
Next we raise it to the power $p$ and get for some p-adic integer $u$,
$$x^p = (\omega+pz)^p = \sum_{n=0}^p \binom{p}{n} \omega^{p-n}(pz)^n = \omega^p + p^2 u = \omega + p^2u$$
Similarly we can recognize $x+1= \omega+1 +pz = -\omega^2 +pz$ and do the same thing for for some p-adic integer $v$,
$$(x+1)^p = (-\omega^2)^p + p^2v = - \omega^2 + p^2v$$
Now we can look at,
$$x^p + 1 - (x+1)^p = \omega + p^2u +  1 + \omega^2 + p^2v = \omega^2+\omega+1 + p^2(u+v) = 0 \mod p^2 $$
A side note, there's nothing special about $p^2$ here, we could have kept raising to higher powers to get $$(x+1)^{p^n} = x^{p^n}+1 \mod p^{n+1}$$
And of course, in the limit this will recover exactly the identity $-\omega^2 = \omega+1$ again in $\mathbb{Q}_p$.
