Proof that $23^{n} - 1$ is divisible by $11$ for all positive integers $n$. I'm having a bit of a problem proving this statement. Maybe someone can point me in the right direction?
Best regards,
 A: HINT:
$$23^{n+1}-1=23(\underbrace{23^n-1})+23-1$$
A: Below I answer you query to Lab about where the proof comes from. As Lab replied, his proof employs  a special case of the following identity. Below I reveal its arithmetical genesis.
$$a^{n+1}-b^{n+1} =\, a(a^n-b^n)+b^n(a-b)$$
That identity shows that  $\ m\mid a^{n+1}-b^{n+1}$ if  $\,m\mid a^n-b^n,\,$  $\,m\mid a-b\ $ (the induction step)
Or, in congruence language $\,a^{n+1}\equiv\, b^{n+1}\,$ if $\,\ a^n\equiv b^n,\ \,$ and $\,\ a\equiv b\pmod m$
Note that this may be viewed as multiplying $\,a^n\equiv b^n\,$ times $\,a\equiv b\,$ using the Product Rule presented below. Indeed, Lab's identity is precisely the identity used in the first proof below.

$\begin{eqnarray}
\rm {\bf Lemma}\ \   &\rm m\ \ |&\rm\ \, X\!-\!x\quad\  \& &&\rm\! m\ |\: Y\!-\!y \ \Rightarrow\ m\:|\!\!&&\rm XY - \: xy\quad {\bf [Divisibility\ Product\ Rule]} \\ \\
\rm {\bf Proof}\ \ \ \ \ &\rm m\ \ |&\rm (X\!-\!\color{#c00}x)\:\color{#0a0}Y\  \,+   &&\rm\, \color{#c00}x\ (\color{#0a0}Y\!-\!y)\ \ \ \ = &&\rm XY - \: xy \\
\\
\rm {\bf Lemma}\ \ &         &\rm\ \, X\equiv x\quad\ \  \& &&\rm\quad\  Y\equiv y \ \ \ \ \Rightarrow\ &&\rm XY\equiv xy \quad {\bf [Congruence\ Product\ Rule]}\\ \\
\rm {\bf Proof}\ \ \ \ \ &0\equiv& \rm (X\!-\!\color{#c00}x)\:\color{#0a0}Y\,\   + &&\rm\, \color{#C00}x\ (\color{#0a0}Y\!-\!y)\ \ \ \ \equiv &&\rm XY - \: xy \\
\end{eqnarray}$ 

Once congruences are known, we may simply quickly apply the Congruence Product Rule
$$ \color{#c00}{23^{\large n}\equiv 1},\ \ \color{#0a0}{23\equiv 1}\,\Rightarrow\, 23^{\large n+1}= \color{#c00}{23^{\large n}}\,\color{#0a0}{23}\equiv \color{#c00}{(1)}\color{#0a0}{(1)}\equiv 1\pmod{11}$$
Thus, with the help of modular language, we see that the induction simplifies to the trivial induction that $\, a\equiv 1\,\Rightarrow\, a^n\equiv 1\,$ (here $\,\color{#0a0}{a = 23}).\,$ This proof can be discovered mechanically, i.e. without any insight or intuition, by using the method of multiplicative telescopy.
A: $$23^n-1=(22+1)^n-1=-1+\sum_{k=0}^n {n \choose k}22^{n-k}1^k$$
Now the only term in the sum that is not divisible by 11 is the one when $n=k$ it means that the reminder is $1^n$,so basically for any number it follows that $$x^n \bmod b=(x\bmod b)^n$$
The other method is math induction
$n=1$
$$23^1-1 =22=0\bmod 11\\23^n-1=0 \bmod 11\\23^{n+1}-1=23^{n+1}-23+22=23(23^n-1)+22$$
Since both $23^n-1$ and $22$ are divisible by 11 it follows that for any n $23^n-1$ is divisible by 11
