Answer
See the proof below.
Work Step by Step
Consider the following statement:
$$a^0+a^1+a^2+a^3+\ldots+a^n=\displaystyle\frac{1-a^{n+1}}{1-a}$$
for all integers $n>0$, $a\neq 1$.
We will use mathematical induction to prove that the statement is true.
$\textbf{1.}$ We have that the statement is true for $n=1$ and for all $a\neq 1$ since
$$a^0+a^1=1+a=\displaystyle\frac{\left(1+a\right)\left(1-a\right)}{1-a}=\displaystyle\frac{1-a^2}{1-a}=\displaystyle\frac{1-a^{1+1}}{1-a}.$$
$\textbf{2.}$ Suppose the statement is true for $n=k$, that is, suppose that
$$a^0+a^1+a^2+a^3+\ldots+a^k=\displaystyle\frac{1-a^{k+1}}{1-a}$$
for all $a\neq 1$.
Then, for any given $a\neq 1$, adding $a^{k+1}$ to both sides of the previous equality gives us that
$$a^0+a^1+a^2+a^3+\ldots+a^k+a^{k+1}=\displaystyle\frac{1-a^{k+1}}{1-a}+a^{k+1}$$
Thus, since
$$\displaystyle\frac{1-a^{k+1}}{1-a}+a^{k+1}=\displaystyle\frac{1-a^{k+1}}{1-a}+\displaystyle\frac{\left(1-a\right)a^{k+1}}{1-a}=\displaystyle\frac{1-a^{k+1}+a^{k+1}-a^{k+2}}{1-a}=$$
$$\displaystyle\frac{1-a^{k+2}}{1-a}=\displaystyle\frac{1-a^{\left(k+1\right)+1}}{1-a}$$
we have that
$$a^0+a^1+a^2+a^3+\ldots+a^k+a^{k+1}=\displaystyle\frac{1-a^{\left(k+1\right)+1}}{1-a}$$
Therefore, the statement holds for $n=k+1$ whenever it is true for $n=k$.
Hence, the mathematical induction principle guarantees that the statement is true for all integers $n>0$ and $a\neq 1$.
