<s>[INST] Prove by induction $\frac{7}{8}+7\left(\frac{1}{8}\right)^2+...+7\left(\frac{1}{8}\right)^n=1-\frac{1}{8^n}$ for every $n \in \mathbb N$ 
$\frac{7}{8}+7\left(\frac{1}{8}\right)^2+...+7\left(\frac{1}{8}\right)^n=1-\frac{1}{8^n}$ for every $n \in \mathbb N$

I first set $n=k$:
$\frac{7}{8}+7\left(\frac{1}{8}\right)^2+...+7\left(\frac{1}{8}\right)^k=1-\frac{1}{8^k}$ 
And then $n=k+1$:
$1-\frac{1}{8^k}+7\left(\frac{1}{8}\right)^{k+1}=1-\frac{1}{8^{k+1}}$ 
And then I tried some Algebra on it:
$1-\frac{1}{8^k}+\left(\frac{7}{8}\right)\left(\frac{7}{8}\right)^k=1-\frac{1}{8^{k+1}}$
But not really sure what to do from here.  Hints?
 [/INST] First comment: you should begin by proving the base case, i.e. for $n=1$.
That being said, for the inductive step:
$$
1-\frac{1}{8^k}+7\frac{1}{8^{k+1}}
=
1-\frac{1}{8^k}+\frac{7}{8}\frac{1}{8^{k}}
=
1-\frac{8}{8}\frac{1}{8^k}+\frac{7}{8}\frac{1}{8^{k}}=
1-\frac{1}{8}\frac{1}{8^k}
=1-\frac{1}{8^{k+1}}
$$
and you can conclude.
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