Proof
Consider the partial sum
\begin{gather*}
S_n = \sum_{k=0}^{n} {x^k} = 1 + x + x^2 + ... + x^n \\
(1-x)S_n = S_n - x S_n = 1 + x + x^2 + ... + x^n - (x + x^2 + ... + x^n + x^{n+1}) = 1 - x^{n+1} \\
\Rightarrow S_n = \frac{1-x^{n+1}}{1-x}
\end{gather*}
and so as \(n \rightarrow \infty \text{,}\)
\begin{gather*}
S_n \rightarrow S = \frac{1}{1-x}
\end{gather*}