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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.

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    Hi, Is there any special type of formula that you're looking for? I'm trying to find analogs such as ${(x)_q}^n$ and ${(x^n)}_q$, and I might try looking for some others, but I'm not sure if you want to restrict the possible formulas in any way.2012-02-15
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    @MattGroff Hi, Matt, I'm specifically looking for an analog for $((r)_q)^n$ to be in the same spirit of $x^n$.2012-02-15

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