Sums $\sum_{k = 0}^n k^t {n \choose k}$ where $t$ is a positive integer

By using Stirling numbers of the second kind we have that: $$ k^t = \sum_{j=0}^{t}j!{t \brace j}\binom{k}{j} $$ hence: $$\sum_{k=0}^n k^t \binom{n}{k} = \sum_{j=0}^{t}{t \brace j}\sum_{k=0}^{n}j!\binom{k}{j}\binom{n}{k}\tag{1}$$ but since: $$\sum_{k=0}^{n}\binom{k}{j}\binom{n}{k} = 2^{n-j}\binom{n}{j}\tag{2}$$ it follows that:

$$\sum_{k=0}^n k^t \binom{n}{k}=2^{n-t}\sum_{j=0}^{t}{t\brace j}\,2^{t-j}\,(n)_j \tag{3}$$

where $(n)_j$ is the falling Pochhammer symbol $(n)_j = n\cdot(n-1)\cdot\ldots\cdot(n-j+1)=j!\binom{n}{j}$.