q-Analogue of the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$. The stirling numbers of the second kind satisfy the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$, where $(x)_k$ is the falling factorial. 
Consider the $q$-analog recursive definition of the stirling numbers, given by
$$
\left\{n\atop k\right\}_q=(k)_q\left\{n-1\atop k\right\}_q+q^{k-1}\left\{n-1\atop k-1\right\}_q.
$$
Why do they satisfy an analog to the standard formula, 
$$
((r)_q)^n=\sum_k\left\{n\atop k\right\}_q(r)_q(r-1)_q\cdots(r-k+1)_q?
$$
Thank you.
 A: The $q$-Stirling numbers defined by your recurrence satisfy the operator equation
$$(xD_q)^n=\sum_k\left\{n\atop k\right\}_qx^kD_q^k\;,\tag{1}$$
where $D_q$ is the $q$-derivative operator defined by $$D_qf(x)=\frac{f(qx)-f(x)}{qx-x}$$ and satisfying $$D_qx^n=\frac{(qx)^n-x^n}{(q-1)x}=\frac{q^n-1}{q-1}x^{n-1}=(n)_qx^{n-1}$$ and $$D_qx=qxD_q\;;$$ for a proof see this answer.
Now $xD_qx^r=x(r)_qx^{r-1}=(r)_qx^r$, so an easy induction shows that $(xD_q)^nx^r=\big((r)_q\big)^nx^r$. Thus, applying $(1)$ to $x^r$ yields
$$\begin{align*}
\Big((r)_q\Big)^nx^r&=(xD_q)^nx^r\\
&=\sum_k\left\{n\atop k\right\}_qx^kD_q^kx^r\\
&=\sum_k\left\{n\atop k\right\}_qx^k(r)_q(r-1)_q\cdots(r-k+1)_qx^{r-k}\\
&=\sum_k\left\{n\atop k\right\}_q(r)_q(r-1)_q\cdots(r-k+1)_qx^r\;.
\end{align*}$$
Now just substitute $x=1$.
A: If you can read German you find an answer in my lecture notes Elementare q-Identitaeten, Chapter 3. (http://homepage.univie.ac.at/johann.cigler/skripten.html).
Your $ \left\{n\atop k\right\}_q$ is the same as my $ q^{\binom{k}{2}} S[n,k]$ 
