Stirling number Combinatorics. Summation .

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$$ \sum_{k=0}^n \left\{ {n\atop k} \right\} *(x)_k = x^n $$ is well known .

What if the k-th term of LHS summation is divided by $q^k$ where $q$ is some positive constant,
What about $$ \sum_{k=0}^n[ \left\{ {n\atop k} \right\} *(x)_k ]/(q^k) = ? $$