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?

You asked for a proof, so here's proof by induction.
Your statement is equivalent to:
$$6 \cdot \sum_{k=1}^n 7^k = 7^{n+1} - 7.$$
This is true for $n=1$:
$$6 \cdot 7 = 7^2 - 7.$$
Now look at the $n+1$ case:
$$6 \cdot \sum_{k=1}^{n+1} 7^k = \left[6 \cdot \sum_{k=1}^{n} 7^k\right] + 6 \cdot 7^{n+1} \\ = 7^{n+1} - 7 + 6 \cdot 7^{n+1} \\= 7 \cdot 7^{n+1} - 7 \\= 7^{n+2} - 7.$$
I invoked the inductive assumption on the second line.
