How can I calculate sum of multichoose multiplied by its argument?

How can I calculate sum of multichoose multiplied by its argument?

I need to calculate sum like this:
$\sum\limits_{k=1}^n k\textstyle\left(\!\!{n\choose k}\!\!\right)$
WolframAlpha is giving nice answer: $n{2n \choose n+1}$ But I don't know
how to prove this result.
Analogous expression for simple binomial coefficients:
$\sum\limits_{k=1}^n k{n\choose k} = n \cdot 2^{n-1}$
can be easily proved by taking derivative of $(1+x)^n$ and setting $x$ to
$1$ after that. But for multichoose I'm dealing with infinite series
$(1-x)^{-n} = \sum\limits_{k=0}^\infty {n-1+k\choose k} x^k$
and solution with setting $x$ to something won't work, I believe.