Complete induction of $10^n \equiv (-1)^n \pmod{11}$ To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.
It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$
For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$
$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.

Why don't you try this: $10 \equiv -1 \ (\text{mod} \ 11) \Longrightarrow (10)^{n} \equiv (-1)^{n} \ (\text{mod} \ 11)$


*

*And for using induction assume it's clear that for $n=1$, the result holds. Now assume that is true for $n=k$. That is $(10)^{k} \equiv (-1)^{k} \ (\text{mod} \ 11)$, and prove it for $n=k+1$.

