Output of discrete-time LTI system guaranteed to be same form as input?
The point is that a complex exponential function is an eigenfunction of an LTI system link. Therefore the frequency is not changing, i.e., the LTI system translates to a (complex-valued) gain $H(j\omega_0)$.

For discrete systems, the eigenfunctions are sampled exponential functions link, i.e., 
$x[n] = A\cdot e^{j\omega_n n}$, where $\omega_0=2\pi f/f_s$ is a normalized frequency. The output of a discrete LTI system is then $y[n]=H(e^{j\omega_n})\cdot A \cdot e^{j\omega_nn}$.

EDIT:
A discrete input signal x[n] can be written as a weighted sum of complex exponentials, i.e., $X(e^{j\omega}) = \sum_{n}x[n]e^{j\omega}$ link. This transformation is in fact a change of basis. Driving such a signal thru an LTI system will weight each exponential function with the complex-valued gain $H(e^{j\omega})$. Thus, no additional frequencies will be introduced.

Note (a bit out of scope, but good to realize): 
When dealing with sampling a contuous-time signal and dealing with finite length signals, then you may observe additional frequencies due to aliasing and windowing.