# Why not use the same “standard” exponentials for both continuous and discrete time

In continuous time the standard exponential signal is usually defined as $$e^{st}, \quad\text{with}\quad s = \sigma+j \omega$$ In discrete time the standard exponential signal is usually defined as $$z^n, \quad\text{with}\quad z=re^{i \theta}$$

Why not define the discrete time signal as $$e^{zn}$$? It seems that $$e^{zn}$$ is also an eigenfunction for discrete LTI systems.

This has to do with the way the Laplace transform and the $$\mathcal{Z}$$-transform are defined:
$$\mathcal{L}\big\{x(t)\big\}=\int_{-\infty}^{\infty}x(t)e^{-st}dt\tag{1}$$
$$\mathcal{Z}\big\{x[n]\big\}=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{2}$$
Of course, the question remains why they are defined that way. The answer lies in the fact that in continuous time we usually have to deal with differentiation and integration, which in the Laplace transform domain correspond to multiplication with $$s$$ and $$1/s$$, respectively. Whereas in discrete time, we usually have to deal with difference equations formulated in terms of time delays and time advances, which in the $$\mathcal{Z}$$-transform domain correspond to multiplications by $$z^{-1}$$ and $$z$$, respectively.
Consequently, the definitions $$(1)$$ and $$(2)$$ lead to transfer functions of linear time-invariant (LTI) systems that are ratios of polynomials in $$s$$ for continuous time, and polynomials in $$z$$ for discrete time. In this way, the choices of $$s$$ and $$z$$ in $$(1)$$ and $$(2)$$ and the corresponding definitions of exponentials as $$e^{st}$$ and $$z^n$$ appear natural.