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.

If $a \equiv b \left(\bmod m \right)$, then $a^n \equiv b^n \left(\bmod m \right)$ where $n \in \mathbb{N}$ and $a,b,m \in \mathbb{Z}$.
The reason being $a^n - b^n$ factors as $(a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + b^{n-1})$ and hence if $m | (a-b)$, $m | (a^n-b^n)$
If you want to prove by induction, first note that
if $a \equiv b \left(\bmod m \right)$ and $c \equiv d \left(\bmod m \right)$, then $a \times c  \equiv b \times d \left(\bmod m \right)$.
This is true since $ac-bd = a(c-d) + d(a-b)$
Hence at the inductive step, if you have that $10^k \equiv (-1)^k \left(\bmod m \right)$, and we have that $10 \equiv (-1) \left(\bmod m \right)$ and from the above result we have
$$10^{k+1} \equiv (-1)^{k+1} \left(\bmod m \right)$$
