Showing that $c_{i}\equiv 0\pmod{p}$ Let the numbers $c_{i}$ be defined by the power series identity $$\frac{1+x+x^{2}+\ldots+x^{p-1}}{(1-x)^{p-1}}= 1+c_{1}x+c_{2}x^{2}+\ldots$$ Show that $c_{i}\equiv 0\pmod{p}$ for all $i\geq 1$.
$\textbf{Solution.}$ First, from the Binomial Theorem we have $$(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!} x^{2} + \frac{n(n+1)(n+2)}{3!}x^{3} + \cdots$$ 
\begin{align*}
(1-x)^{-(p-1)} &= 1 + (p-1)x + \frac{(p-1)p}{2!} x^{2} + \frac{(p-1)p(p+1)}{3!}x^{3} + \cdots \\ 
&= 1 + \binom{p-1}{1} x + \binom{p}{2}x^{2} + \binom{p+1}{3}x^{3} + \binom{p+2}{4}x^{4} + \cdots \\
&= 1+\binom{p-1}{1}x + \binom{p}{2}x^{2} + \left[\binom{p}{2}+\binom{p}{3}\right]x^{3}+\left[\binom{p+1}{3}+\binom{p+1}{4}\right]x^{4} +...\\
&=1+\binom{p-1}{1}x + \binom{p}{2}x^{2} + \left[\binom{p}{2}+\binom{p}{3}\right]x^{3}+\left[\binom{p}{2}+2\binom{p}{3}+\binom{p}{4}\right]x^{4} +...
\end{align*}
where $\binom{p+1}{3}x^{3}=\Bigl\{\binom{p}{2}+\binom{p}{3}\Bigr\}x^{3}$ and $\binom{p}{4}x^{4}=\Bigl\{\binom{p}{2}+2\binom{p}{3}+\binom{p}{4}\Bigr\}x^{4}$ follows from using $\textit{Pascal's Identity.}$ Now we have $$\frac{1+x+x^{2}+\ldots+x^{p-1}}{(1-x)^{p-1}}=(1+x+x^{2}+\ldots)\cdot (1-x)^{-(p-1)} =1+c_{1}x+c_{2}x^{2}+\ldots$$ Using the binomial expansion of $(1-x)^{-(p-1)}$ and comparing the coefficients, we get 
\begin{align*}
c_{1} &= 1 + \binom{p-1}{1} =p\\
c_{2} &= \binom{p}{2} + \binom{p-1}{1} + 1 = \binom{p}{2} + c_{1}\\
c_{3} &= \binom{p+1}{3} + \binom{p}{2} + \binom{p-1}{1} + 1 =\binom{p}{2}+\binom{p}{3}+c_{2}\\
c_{4} &=\binom{p+2}{4}+c_{3} = \binom{p}{2}+2\binom{p}{3}+\binom{p}{4}+c_{3}
\end{align*}
In general $c_{i}=\binom{p+i-2}{i}+c_{i-1}$. Now by we know that $\binom{p}{i}\equiv 0\pmod{p}$ whenever $0<i<p$. 


*

*So I am able to observe that $c_{1},c_{2},c_{3},c_{4}$ are $0\pmod{p}$. But still my argument is not correct. Since $c_{i}\neq\binom{p+i-2}{p}+c_{i-1}$ if $i\geq p$. A nice solution would be helpful. 


In $R:=\mathbb{F}_p[\![x]\!]$, $(1-x)^p=1-x^p$.  Because $1-x$ is an invertible element of $R$ (with inverse $1+x+x^2+\ldots$), $$\frac{1-x^p}{(1-x)^p}=1$$ in $R$.  Therefore, [...].  The rest is up to you.
