Proving the geometric series $\sum_{i=0}^n r^i = \frac{1-r^{n+1}}{1-r}$ by induction for $n\geq 1$ Let $P(n)$ be the statement 
$$
P(n) : \sum\limits^{n}_{i=0}r^i = \dfrac{1-r^{1+n}}{1-r}\text{ for all }n \in \mathbb{N}\text{.}
$$
I am stuck at the base case:
$$P(1):1 + r = \dfrac{1-r^2}{1-r}\text{.}$$

I am stuck as to how I can show $P(1)$ is true.
Note: First note that you are really summing the geometric series ($a,r\in\mathbb{R}, a\neq 0, r\neq 1$)
$$
a+ar+ar^2+\cdots+ar^n=a\frac{r^{n+1}-1}{r-1}=a\frac{1-r^{n+1}}{1-r}
$$
with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.

Claim: Let $a$ and $r$ be real numbers with $a\neq 0$ and $r\neq 1$. Prove that for each integer $n\geq 1$ that
$$
a+ar+ar^2+\cdots+ar^n=a\frac{r^{n+1}-1}{r-1}.
$$
Proof. Fix $r\in\mathbb{R}, r\neq 1$, and let $S(n)$ denote the following statement for $n\geq 1$:
$$
S(n) : 1+r+r^2+\cdots+r^n=\frac{r^{n+1}-1}{r-1}.
$$
In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $n\geq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $r\neq 0$ without loss of generality. 

Base step ($n=1$): The statement $S(1)$ says that $1+r=\frac{r^2-1}{r-1}$, which is true since $r^2-1=(r+1)(r-1)$ and $r\neq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=\frac{r-1}{r-1}$.]

Inductive step ($S(k)\to S(k+1)$): Fix some $k\geq 1$ and suppose that $S(k)$ is true; that is, assume that
$$
S(k) : 1+r+r^2+\cdots+r^k=\frac{r^{k+1}-1}{r-1}
$$
is true. We must show that $S(k+1)$ follows where
$$
S(k+1) : 1+r+r^2+\cdots+r^k+r^{k+1}=\frac{r^{k+2}-1}{r-1}.
$$
Starting with the left-hand side of $S(k+1)$,
\begin{align}
\text{LHS} &amp;= \color{red}{1+r+r^2+\cdots+r^k}+r^{k+1}\tag{by definition}\\[1em]
&amp;= \color{red}{\frac{r^{k+1}-1}{r-1}}+r^{k+1}\tag{by $S(k)$}\\[1em]
&amp;= \frac{r^{k+1}-1}{r-1}+\frac{r^{k+1}(r-1)}{r-1}\tag{common denom.}\\[1em]
&amp;= \frac{r^{k+1}-1+r^{k+1}(r-1)}{r-1}\tag{combine like terms}\\[1em]
&amp;= \frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}\tag{expand}\\[1em]
&amp;= \frac{r^{k+2}-1}{r-1},\tag{simplify}
\end{align}
we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)\to S(k+1)$, completing the inductive step. 

By mathematical induction, for all $n\geq 1, S(n)$ holds true. $\blacksquare$ 



Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $\sum_{i=0}^n 2^i=2^{n+1}-1$. One can easily set $r=2$ in what we just proved to see that this result is true.
$$1+r=\frac{1-r^2}{1-r}$$should lead you to check if $$(1+r)(1-r)=1-r^2.$$
I don't need to tell you more.