Solve Proof by Induction with 2 variables I am not sure how to solve a proof by induction using $2$ variables. Do I solve for $n$ first and then $q$, or vice versa? Can you please give me a hint? Thanks.

Prove by induction on $n$ that for any real number $q>1$ and integer $n\ge 0$: $$1+q+q^2+\cdots+q^{n-1}+q^n=\frac{q^{n+1}-1}{q-1}\;.$$

 A: In order to prove by induction on $n$, the "standard procedure" is to prove for a base case, assume for $n = k$ for some $k \in \Bbb N$ and then to show that the argument also holds for $n = k+1$.
Base Case:
Since we are given that $n \geq 0$, our lowest possible value for $n$ is $0$, so let $n=0$. Then,
$$1 = \frac{q-1}{q-1}$$
which is true, so the statement is true for $n = 0$. Your confusion here is that if $n=0$, the summation is simply
$$\sum_{k=0}^0 q^k = q^0 = 1.$$
Induction hypothesis:
Since we have shown that the statement holds for at least one $n\geq0$, let's assume that it holds for some $k\in \Bbb N$. Then 
$$1 + q + q^2 + \cdots + q^{k-1} + q^k = \frac{q^{k+1} - 1}{q -1}.$$
Proof for $n = k+1$:
Now let's show that the statement holds for $n = k+1;$
If we let $n = k+1$ then we have that 
\begin{align}
1 + q + q^2 + \cdots + q^k + q^{k+1} &= \frac{q^{(k+1) +1} -1}{q-1}\\
&=\frac{q^{k+2} -1}{q-1}.
\end{align}
Notice that, up until the $q^{k+1}$ term, we actually have what we had in our inductive hypothesis. So we can write
$$\frac{q^{k+1} - 1}{q -1} + q^{k+1} = \frac{q^{k+2} -1}{q-1}.$$
Placing the LHS over a common denominator, we have that 
\begin{align}
\frac{q^{k+1} - 1 + q^{k+1}(q-1)}{q -1} &= \frac{q^{k+1} - 1 + q^{k+2}-q^{k+1}}{q -1}\\
&=\frac{q^{k+2} -1}{q-1},
\end{align}
as desired.
