I am following Warren P. Johnson's "An Introduction to q-analysis". We are supposed to prove $$\binom{n+1}{k+1}_q=\sum_{m=k}^{n}q^{m-k}\binom{m}{k}_q$$ Here is my attempt.
We start off by using the $q$-analogue of Pascal's identity, which is $$\binom{n+1}{k}_q=\binom{n}{k}_q + q^{n-k+1}\binom{n}{k-1}_q$$
Setting $n=n-1$ grants \begin{align} \binom{n}{k}_q = \binom{n-1}{k}_q + q^{n-k}\binom{n-1}{k-1}_q \end{align}
Substitute this back into the equation we started with, repeat this for (say) $m$ times to get
\begin{align} \binom{n+1}{k}_q = \binom{n-m}{k}_q + \sum_{i=0}^{m}q^{n-k-i+1}\binom{n-i}{k-1}_q \end{align}
How do I complete the proof? I understand that I have to choose the "right" value for $m$ and then set $k=k+1$.
The result to prove is
$$ \binom{n+1}{k+1}_q=\sum_{m=k}^{n}q^{m-k}\binom{m}{k}_q. \tag1 $$
The $q$-analog of Pascal's identity is
$$ \binom{n+1}{k}_q=\binom{n}{k}_q + q^{n-k+1}\binom{n}{k-1}_q. \tag2 $$
Replace $k$ with $k+1$ to get the identity
$$ \binom{n+1}{k+1}_q=\binom{n}{k+1}_q + q^{n-k}\binom{n}{k}_q. \tag3 $$
Move the $\binom{n}{k+1}_q$ to the other side to get
$$ \binom{n+1}{k+1}_q - \binom{n}{k+1}_q = q^{n-k}\binom{n}{k}_q. \tag4 $$
This implies that equation $(1)$ is a telescoping sum.
Notice that when $n=k-1$ then $\binom{n+1}{k+1}_q = 0.$
The result follows.