With the discrete step function
$$ u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \\ \end{cases} $$
And the output $y[n]$ defined as a discrete convolution of the input $x[n]$ with the impulse response $h[n]$:
$$ y[n] = \sum\limits_{k=-\infty}^{\infty} h[k] x[n-k] $$
If the input is defined as:
$$ x[n] = e^{j\omega n} u[n] $$
What is the output?
My textbook says the answer is:
$$ y[n] = \begin{cases} 0, & n < 0 \\ \left(\sum\limits_{k=0}^n h[k] e^{-j\omega k}\right) e^{j \omega n}, & n \ge 0 \\ \end{cases} $$
I don't see how the book arrives at this answer. My work is:
$$ u[n-k] = \begin{cases} 1, & n \ge k \\ 0, & n < k \\ \end{cases} $$
\begin{align*} y[n] &= \sum\limits_{k=-\infty}^{\infty} h[k] x[n-k] \\ &= \sum\limits_{k=-\infty}^{\infty} h[k] e^{j\omega (n-k)} u[n-k] \\ &= \sum\limits_{k=-\infty}^{n} h[k] e^{j\omega (n-k)} \\ &= \sum\limits_{k=-\infty}^{n} h[k] e^{-j\omega k} e^{j\omega n} \\ \end{align*}
My work seems correct, but it clearly doesn't match the text answer. I see why the input is zero for $n < 0$, because $u[n]$ will be zero, but I don't see why the output will be zero for $n < 0$.
Your calculations are indeed correct. However, the textbook answer is based on the assumption that the impulse response $h[k]$ is causal in that $h[k]=0$ for $k<0$ - otherwise the filter output $y[n]$ would be dependent on future values of the input $x[n]$.