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 $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &amp;= k*11\\
(-1)*(-1)^n - 10*(10)^n &amp;= k*11
\end{align*}$$
But I don't get the next step.
You are not setting up your induction very well. You should not start with the equality you want to establish, namely that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$. Instead, you should start with the Induction Hypothesis, which is that $(-1)^n - 10^n$ is a multiple of $11$.

So: the Inductive Step is to show that if $(-1)^n - 10^n$ is a multiple of $11$, then $(-1)^{n+1} - 10^{n+1}$ is also a multiple of $11$.

Let's write out our Induction Hypothesis: it says that
$$\text{There exists an integer }k\text{ such that }(-1)^n - 10^n = 11k.$$

What we want to prove is that:
$$\text{there exists an integer }\ell\text{ such that }(-1)^{n+1}-10^{n+1}=11\ell.$$
(Note that the multiple may be different, that's why I used a different letter).

So now we can try manipulating the expression we want. One possibility is to use the following identity:
$$a^{n+1}-b^{n+1} = (a-b)(a^n + a^{n-1}b + a^{n-2}b^2 + \cdots + ab^{n-1}+b^n),$$
if you already know this identity. 

So we have, with $a=-1$ and $b=10$,
$$
(-1)^{n+1} - 10^{n+1} = \Bigl( (-1) - 10\Bigr)\Bigl( (-1)^n + (-1)^{n-1}(10) + \cdots + (-1)10^{n-1} + 10^n\Bigr).$$
Now notice that you don't even need to use the induction hypothesis to conclude that $(-1)^{n+1}-10^{n+1}$ is a multiple of $11$ (as could be seen in mac's answer).

If you don't know the identity, then you can perform some purely algebraic manipulations. E.g.,
$$\begin{align*}
(-1)^{n+1} - 10^{n+1} &amp;= -1\left( (-1)^n + 10^{n+1}\right)\\
 &amp;= -\left( (-1)^n -10^n + 10^n + 10^{n+1}\right)\\
&amp;= -\left( \Bigl((-1)^n - 10^n\Bigr) + 10^n\Bigl(1 + 10\Bigr)\right)\\
&amp;= -\left( 11k + 10^n(11)\right) &amp;\quad&amp;\text{(by the induction hypothesis)}\\
&amp;= -\left( 11(k+10^n)\right)\\
&amp;= 11\left( -(k+10^n)\right),
\end{align*}$$
which gives that $(-1)^{n+1} - 10^{n+1}$ is a multiple of $11$, as desired, from the assumption that $(-1)^n - 10^n$ is a multiple of $11$.



But easier still is to use the following property of congruences:

Proposition. Let $a,b,c,d,k$ be integers. If 
$$a\equiv b\pmod{k}\qquad\text{and}\qquad c\equiv d\pmod{k}$$
then $ac\equiv bd\pmod{k}$.

Proof. Since $a\equiv b\pmod{k}$, then $k|a-b$, so $k$ divides any multiple of $a-b$; for example, $k|(a-b)c = ac-bc$. Since $k$ divides $ac-bc$, then $ac\equiv bc\pmod{k}$.

Since $c\equiv d\pmod{k}$, then $k|c-d$, so $k|(c-d)b = cb-db$, hence $bc\equiv bd\pmod{k}$. 

Since $ac\equiv bc\pmod{k}$ and $bc\equiv bd\pmod{k}$, then $ac\equiv bd\pmod{k}$. QED

Corollary. If $a_1\equiv b_1\pmod{k}$, $a_2\equiv b_2\pmod{k},\ldots, a_n\equiv b_n\pmod{k}$, then 
$$a_1\cdots a_n\equiv b_1\cdots b_n\pmod{k}.$$

Proof. Induction on $n$. QED

(This is where you would want to use induction, rather than the specific case you are looking at).

Corollary. If $a\equiv b\pmod{k}$, then for all positive integers $n$, $a^n\equiv b^n\pmod{k}$.

Proof. Apply previous corollary with $a_i=a$ and $b_i=b$ for all $i$. QED
Direct proof: $10^n=(11-1)^n=11(\dots)+(-1)^n$