\begin{align*} \mu & = E[X] = \sum_{x=1}^{\infty} {x(1-p)^{x-1}p}\\ & = p \sum_{x=1}^{\infty} {x(1-p)^{x-1}}\\ & = p \frac{1}{(1-(1-p))^2}\\ & = p \frac{1}{p^2} = \frac{1}{p} \end{align*}