Q:

Derivation of discrete-time steady-state and transient response

My discrete-time signals textbook (Oppenheim) jumps through the derivation of steady-state and transient LTI system response to a complex exponential input-
$$y[n]=y_{SS}[n]+y_t[n]$$
It says, given input \$x[n]=e^{j\omega n}u[n]\$ and impulse response \$h[n]\$, the system response can be found using convolution as:
$$
y[n] = \left\{\begin{aligned}
&0 &&: n < 0\\
&e^{j\omega n}\sum_{k=0}^{n}h[k]e^{-j\omega k} &&: n\ge 0
\end{aligned}
\right.$$
But where did the \$n\$ as the upper sum limit come from? When I try to do this myself, I end up with:
$$
y[n]=x[n]\ast h[n]=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\\
=\sum_{k=-\infty}^{\infty}h[k]e^{j\omega(n-k)}u[n-k]\\
=\sum_{k=0}^{\infty}h[k]e^{j\omega n}e^{-j\omega k}\\
=e^{j\omega n}\sum_{k=0}^{\infty}h[k]e^{-j\omega k}
$$
Where did I go wrong?

A:

Given: the system is LTI and causal.
$$\Rightarrow y[n]=h[n]\ast x[n]=\sum_{k=-\infty}^{\infty}h[k]e^{j\omega (n-k)}u[n-k]\\~\\
\because \textrm{ causality implies }h[k]=0\ \forall k<0\\
\because u[n-k]=0\ \forall k>n\\~\\
\therefore \sum_{k=-\infty}^{\infty}h[k]e^{j\omega (n-k)}u[n-k]=e^{j\omega n}\sum_{k=0}^{n}h[k]e^{-j\omega k}
$$
Combining that with the fact that the output is causal-
$$
y[n] = \left\{\begin{aligned}
&0 &&: n < 0\\
&e^{j\omega n}\sum_{k=0}^{n}h[k]e^{-j\omega k} &&: n\ge 0
\end{aligned}
\right.
$$

