The Z transform: why $z^{-n}$ and not $z^n$? The Z-transform for a discrete signal is defined as $X(z) = \sum_{n = -\infty}^{\infty} x[n]z^{-n}$. I was wondering why we invert the exponent of $z$, rather than define it as $X(z) = \sum_{n = -\infty}^{\infty} x[n]z^{n}$, which to my mind seems like a more natural definition.
I've not been able to find an answer for this anywhere. Both definitions seem to give pretty much the same theory (but different domains of convergence in applications).
 A: The Wikipedia article Z-Transform explains that 

In geophysics, the usual definition for the $Z$-transform is a power series in $z$ as opposed to $z^{−1}$.

and continues

The two definitions are equivalent; however, the difference results in a number of changes. For example, the location of zeros and poles move from inside the unit circle using one definition, to outside the unit circle using the other definition

Thus it is a matter of convention or preference. Something analogous applies to the Laplace transform and Fourier transform, especially for Fourier transform with differing factors.
A: I'll try to explain why $z$ transform is easy to use in control systems and not power series. Consider a system with block diagram 

It has the system functional
$$\dfrac{Y}{X}=\mathcal{H(R)}=\dfrac{1}{1-\mathcal{R-R^2}}=\sum\limits_{n=-\infty}^{\infty}h[n]\mathcal{R}^{\color{red}{n}}$$
You could try long division or partial fractions and get the unit sample response $h[n]$. They all work pretty good if all you want is just the unit sample response. The only drawback is you can't see the poles directly in above form. So we engineers, being lazy, replace $\mathcal{R}$ by $\dfrac{1}{z}$ so that the roots of the quadratic in $z$ in the denominator give the poles. After this replacement, we call it the  system function  :
$$H(z) = \dfrac{Y(z)}{X(z)}=\dfrac{z^2}{z^2-z-1}=\sum\limits_{n=-\infty}^{\infty}h[n]z^{\color{red}{-n}}$$
