Consider the signal $ x[n]=a^{n}u[n] $
$ X(z)=\sum_{n=-\infty}^{\infty}a^{n}u[n]z^{-n}=\sum_{n=0}^{\infty}(az^{-1})^{n} $
for convergence,
$ \left| az^{-1}\right|<1 $, or equivalently $ \left| z\right|>\left| a\right| $
$ X(z)=\sum_{n=0}^{\infty}(az^{-1})^{n}=\frac{1}{1-az^{-1}}=\frac{z}{z-a} $
