<s>[INST] How continue with this proof? I'm trying to proof the following statement:
$7+6 \cdot (7+7^1+7^2+\dots+7^n) = 7^{n+1}$
I try the next:
$n=k$
$7+6 \cdot (7+7^1+7^2+\dots+7^k) = 7^{k+1}$
$7+6 \cdot (7+7^1+7^2+\dots+7^k)-7 = 7^{k+1}-7$
$\frac{6 \cdot (7+7^1+7^2+\dots+7^k)}{6} = \frac{7^{k+1}-7}{6}$
$7+7^1+7^2+\dots+7^k = \frac{7^{k+1}-7}{6}$
$n=k+1$
$7+7^1+7^2+\dots+7^k+7^{k+1} = \frac{7^{k+2}-7}{6}$
Proof:
$\frac{7^{k+1}-7}{6}+7^{k+1} = \frac{7^{k+2}-7}{6}$
$\frac{7^{k+1}-7+6\cdot7^{k+1}}{6}$
Any idea?
 [/INST] It seems you want to demonstrate by induction that
$$ 
7 + 6 \cdot \left( {7 + 7^{\,2\,}  +  \cdots  + 7^{\,n\,} } \right) = 7^{\,n + 1\,}  
$$
Now, that's true for $n=1$ 
$$ 
\begin{gathered} 
  n = 1 \hfill \\ 
  7 + 6 \cdot \left( 7 \right) = 7 \cdot \left( {6 + 1} \right) = 7^{\,2\,}  \hfill \\  
\end{gathered}  
$$
and we want to demonstrate that if it is valid for $n$ it is also valid for $n+1$.
So we take our identity, multiply both sides by $7$, do some massage 
$$ 
\begin{gathered} 
  n \to n + 1 \hfill \\ 
  7 + 6 \cdot \left( {7 + 7^{\,2\,}  +  \cdots  + 7^{\,n\,} } \right) = 7^{\,n + 1\,}  \hfill \\ 
  7 \cdot \left( {7 + 6 \cdot \left( {7 + 7^{\,2\,}  +  \cdots  + 7^{\,n\,} } \right)} \right) = 7 \cdot 7^{\,n + 1\,}  = 7^{\,n + 2\,}  \hfill \\ 
  \left( {7 \cdot 7 + 6 \cdot 7 \cdot \left( {7 + 7^{\,2\,}  +  \cdots  + 7^{\,n\,} } \right)} \right) = 7^{\,n + 2\,}  \hfill \\ 
  \left( {7 \cdot \left( {6 + 1} \right) + 6 \cdot 7 \cdot \left( {7 + 7^{\,2\,}  +  \cdots  + 7^{\,n\,} } \right)} \right) = 7^{\,n + 2\,}  \hfill \\ 
  \left( {7 + 6 \cdot 7 + 6 \cdot \left( {7^{\,2\,}  + 7^{\,3\,}  +  \cdots  + 7^{\,n + 1\,} } \right)} \right) = 7^{\,n + 2\,}  \hfill \\ 
  \left( {7 + 6 \cdot \left( {7 + 7^{\,2\,}  + 7^{\,3\,}  +  \cdots  + 7^{\,n + 1\,} } \right)} \right) = 7^{\,n + 2\,}  \hfill \\  
\end{gathered}  
$$
and get the demonstration   
</s>