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.