Show $ 1 + a + a^2 + a^3 + \ldots + a^n = \frac{a^{n+1}-1}{a-1}$ by induction How can we show by mathematical induction that the following holds for $ n \ge 0$ and $a \ne 1$?
$$ 1 + a + a^2 + a^3 + \ldots + a^n = \frac{a^{n+1}-1}{a-1}$$
I understand the principle of mathematical induction, but I've no idea how to apply it here. 
I know in general I have to prove it for $n = 1$ and then assume $n = k$ is correct. Then prove $n = k+1$ is true.
But what about $a$?
I've watched a load of YouTube videos on the subject, and I've read a few questions here but it's not helping. The videos make sense while I'm watching them, but I don't know how to apply it. 
This question appeared on my discrete math exam last week. I did not do well. I think I'm missing something fundamental in my understanding of this subject.
 A: If $n=0$, we must prove $1=\frac{a-1}{a-1}$ which is trivial. 
Assume that it holds for $n=k$, that is: $$1+a+...+a^k=\frac{a^{k+1}-1}{a-1}$$ We must prove that it holds for $n=k+1$ or in other words that
$$1+a+...+a^k+a^{k+1}=\frac{a^{k+2}-1}{a-1}$$
But this is simple to prove with our assumption: $$1+a+...+a^k+a^{k+1}=\frac{a^{k+1}-1}{a-1}+a^{k+1}=\frac{a^{k+1}-1+a^{k+2}-a^{k+1}}{a-1}=\frac{a^{k+2}-1}{a-1}$$
A: Let $P(n)$ be the statement that $1+a+...+a^n=\frac{a^{n+1}-1}{a-1}$. Since $1=\frac{a-1}{a-1}$, $P(0)$ is true.
Suppose $P(k)$ is true for some non-negative integer $k$. Then $1+a+...+a^k=\frac{a^{k+1}-1}{a-1}$, so that $1+a+...+a^{k+1}=\frac{a^{k+1}-1}{a-1}+a^{k+1}=\frac{a^{k+1}-1}{a-1}+\frac{a^{k+2}-a^{k+1}}{a-1}=\frac{a^{k+2}-1}{a-1}$, so that $P(k+1)$ is true.
Hence we have $P(n)$ for all non-negative integers $n$.
Note that in this example, you are asked to start at $n=0$ and not $n=1$ and thus prove the statement for all non-negative integers and not just positive integers. Also note that here $a\neq 1$ is fixed and we are doing induction on $n$ and not on $a$. 
