Generalization of a minor LTE lemma The problem asks to prove that for every odd prime $p$ and integers $x,n$ (where $n>0$ , $x\neq1$) such that $p\mid x-1$ , $\nu_p(n)=\alpha$, one has the following:
$$\frac{x^n-1}{x-1}\overset{p^{\alpha+1}}{\equiv\hspace{-2 pt}\equiv\hspace{-1 pt}\equiv}\hspace{2 pt}n$$
I have almost no idea how to prove this one. All I could do was some reformation as the following:
$$p^{\alpha+1}\mid\frac{x^n-1}{x-1}-n=\displaystyle\sum_{i=0}^{n-1}x^i-n=\displaystyle\sum_{i=0}^{n-1}(x^i-1)=\displaystyle\sum_{i=1}^{n-1}(x^i-1)=(x-1)\displaystyle\sum_{i=0}^{n-2}x^i(n-i-1)$$
But I couldn't take this any further. This is clearly a stronger version of the well-known LTE lemma. I tried to tackle the problem with the ideas used in the proof of LTE as well but got to nowhere.
Any help would be appreciated!
 A: I don't know of any way to directly continue from where you left off (when I tried, I just got $\sum_{i=0}^{n-1}x^i - n \equiv \sum_{i=0}^{n-1}x^i - nx^{n-1} \pmod{p^{\alpha+1}}$). Instead, using what you've already determined, induction can be used on $\alpha$ to prove
$$p^{\alpha + 1} \mid \sum_{i=0}^{n-1}x^i - n \; \; \to \; \; \sum_{i=0}^{n-1}x^i - n \equiv 0 \pmod{p^{\alpha + 1}} \tag{1}\label{eq1A}$$
Note that $\nu_p(n) = \alpha$ means for some positive integer $b$ that
$$n = bp^{\alpha}, \; \; p \nmid b \tag{2}\label{eq2A}$$
Also, $p \mid x - 1$ means for some non-zero integer $c$ that
$$x = cp + 1 \tag{3}\label{eq3A}$$
For $\alpha = 0$, since $n = b$ and $x \equiv 1 \pmod{p} \; \to \; x^i \equiv 1 \pmod{p}$, we get
$$\sum_{i=0}^{b-1}x^i - b \equiv \sum_{i=0}^{b-1}1 - b \equiv b - b  \equiv 0 \pmod{p} \tag{4}\label{eq4A}$$
Assume \eqref{eq1A} holds for $\alpha = k$ for some $k \ge 0$, so $n = bp^k$ from \eqref{eq2A}. This means, for some integer $m$, that
$$\sum_{i=0}^{bp^k-1}x^i = mp^{k+1} + bp^k \tag{5}\label{eq5A}$$
We next get
$$\begin{equation}\begin{aligned}
\sum_{i=0}^{bp^{k+1}-1}x^i - bp^{k+1} & = \sum_{j=0}^{p-1}x^{j(bp^k)}\left(\sum_{i=0}^{bp^k-1}x^i\right) - bp^{k+1} \\
& = \sum_{j=0}^{p-1}x^{j(bp^k)}\left(mp^{k+1} + bp^k\right) - bp^{k+1} \\
& = mp^{k+1}\sum_{j=0}^{p-1}x^{j(bp^k)} + bp^k\sum_{j=0}^{p-1}x^{j(bp^k)} - bp^{k+1} \\
& = mp^{k+1}\sum_{j=0}^{p-1}x^{j(bp^k)} + bp^k\left(\sum_{j=0}^{p-1}x^{j(bp^k)} - p\right)
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Also, using \eqref{eq3A} and the binomial theorem, we have that
$$\begin{equation}\begin{aligned}
\sum_{j=0}^{p-1}x^{j(bp^k)} & \equiv \sum_{j=0}^{p-1}(cp + 1)^{j(bp^k)} \\
& \equiv \sum_{j=0}^{p-1}\left(\sum_{i=0}^{j(bp^k)}\binom{j(bp^k)}{i}(cp)^{i}\right) \\
& \equiv \sum_{j=0}^{p-1}(1 + j(bp^k)(cp)) \\
& \equiv \sum_{j=0}^{p-1}1 + bcp^{k+1}\sum_{j=0}^{p-1}j \\
& \equiv p + bcp^{k+1}\left(\frac{p(p-1)}{2}\right) \\
& \equiv p + bcp^{k+2}\left(\frac{p-1}{2}\right) \\
& \equiv p \pmod{p^2}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Using this in \eqref{eq6A} gives that both terms in the last line are congruent to $0$ modulo $p^{k+2}$, resulting in
$$\sum_{i=0}^{bp^{k+1}-1}x^i - bp^{k+1} \equiv 0 \pmod{p^{k+2}} \tag{8}\label{eq8A}$$
This proves that \eqref{eq1A} also holds for $\alpha = k + 1$. Thus, by induction, \eqref{eq1A} is true for all $\alpha \ge 0$.
