Modulo equation : $\frac{n^{k+1}-1}{n-1} \equiv a{\pmod p}$ Can we have directly answer for this question :
$$\frac{n^{k+1}-1}{n-1} \equiv a{\pmod p}$$ 
(p is a prime and k,n is a fixed number)
My question is : with fixed number n, k and p, can we know value a ? 
Thanks :)
 A: Let's us assume  $n≠1$,else the L.H.S. will be undefined, 
also assume $n$ is any integer and integer $k≥-1$.  
$(1)$If $k=-1,a\equiv 0{\pmod p}$ .
$(2)$If $k=0,a\equiv 1{\pmod p}$ .
$(3)$If $k>0$,
$(A)$If $p\mid (n-1)$ i.e., $n=1+ap$ for some integer $a$
Then, $n^{k+1}-1=(1+ap)^{k+1}-1=(k+1)ap+^{k+1}C_2(ap)^2+...$
So,
$\frac{n^{k+1}-1}{n-1}$
 $$=\frac{(k+1)ap+^{k+1}C_2(ap)^2+...}{ap}\equiv k+1 \pmod p \implies a \equiv k+1 \pmod p $$
$(B)$If $p∤(n-1)$ i.e., $(n-1,p)=1$, 
(i)If $p\mid n,$
$$ \frac{n^{k+1}-1}{n-1} \equiv 1 \pmod p\implies  a \equiv 1 \pmod p$$
(ii)else $(p,n)=1$, let $ord_pn=d$ which is clearly $>1$.
If $k+1=q\cdot d+ r$ where $0≤r<d$,$n^{k+1}\equiv n^r \pmod p$
$$\implies n^{k+1}-1\equiv n^r-1 \pmod p$$
$$\implies \frac{n^{k+1}-1}{n-1}\equiv \frac{n^{r}-1}{n-1}\pmod p$$ as $(n-1,p)=1$
$$\implies a \equiv \frac{n^{r}-1}{n-1} \pmod p$$
$\equiv 0$ if $r=0$,
$\equiv(n^{r-1}+...+n+1) $ otherwise.
If $n\equiv m$ where $m$ can be in $(1,p-1)$ or in $(-\frac{p}{2}, \frac{p}{2})$, 
$n^t$ (where $t$ is any natural number) can be replaced with $m^t$ for the ease of calculation.
