# Elementary number theory on congruences

Let the numbers $c_i$ be defined by the power series identity $(1+ x + x^2+ ... + x^n)/(1-x)^{p-1} = 1 + c_1x + c_2x^2+...$. Show that $c_i \equiv 0 \bmod{p}$ for all $i \geq 1$

Any help on starting out, please? Thank u!

Hint: Since for all suitable $x$ we have $1+x+\cdots+x^n=\frac{1-x^{n+1}}{1-x}$, we are looking at $$\frac{1-x^{n+1}}{(1-x)^p}.$$ Now write down the power series expansion of $\frac{1}{(1-x)^p}$.
Remark: Note however that in general we do not have $c_i\equiv 0\pmod{p}$. Try for example $\frac{1+x}{(1-x)^2}$ (so $n=1$, $p=3$). But for suitable values of $n$, the result holds.
• $1+px+(p^2+p)x^2/2+(p^3+3p^2+2p)x^3/6+...$ right? – phoenix Mar 6 '14 at 5:03