How to prove double factorial sum: $\pi\frac{(-1)^{n+1}}{2^{2n-3}}\sum_{k=0}^{n-1}(-1)^k{2n\choose k}(n-k)=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$? 
$$\pi\cdot\frac{(-1)^{n+1}}{2^{2n-3}}\sum_{k=0}^{n-1}(-1)^k{2n\choose k}(n-k)=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2},\ \ \ \ n\in\Bbb{N}^+\tag1$$

I have a feeling that $(1)$ might be true. Contour integration yielded the LHS while the RHS is given in the answer key. To prove this result without induction, I first I distributed the sum, and showed that
$$\begin{align} n\sum_{k=0}^{n-1}(-1)^k{2n\choose k}&=n\sum_{k=0}^{n-1}(-1)^k{2n\choose k}{n-k-1\choose n-k-1}\\&=n\sum_{k=0}^{n-1}(-1)^{k+n-k-1}{2n\choose k}{-1\choose n-k-1}\\ &=n(-1)^{n-1}{2n-1\choose n-1}.\end{align}$$
I am not sure what to do with 
$$\sum_{k=1}^{n-1}k(-1)^k{2n\choose k}\tag2$$
since this has the additional $k$. I thought about using some definite integral substitutions for $k$ but those did not get me anywhere. A hint to solving $(2)$ ought to suffice though complete answers are also welcome.
EDIT. The original equality in $(1)$ does not hold. I had messed up the denominator. The correct identity is

$$\pi\cdot\frac{(-1)^{n+1}}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^k{2n\choose k}(n-k)=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2},\ \ \ \ n\in\Bbb{N}^+\tag{1'}.$$

Thanks to Felix Marin for pointing this out!
 A: We have
$$ k\binom{2n}{k} = \frac{(2n)!}{(k-1)!(2n-k)!} = 2n\binom{2n-1}{k-1} $$
hence it follows that:
$$ A(n) = \sum_{k=1}^{n-1} k(-1)^k \binom{2n}{k} = 2n\sum_{k=1}^{n-1}(-1)^k \binom{2n-1}{k-1} = -2n\sum_{k=0}^{n-2}(-1)^k \binom{2n-1}{k} $$
and you already know how to deal with the last sum by exploiting symmetry.

An alternative comes from experience: the RHS of your identity is clearly related with integrals of the form $\int_{0}^{\pi/2}\sin(x)^{2n}\,dx$. So we may start considering that
$$(e^{it}-e^{-it})^{2n} = \sum_{k=0}^{2n}\binom{2n}{k}e^{(2n-k)it}e^{-kit}(-1)^k $$
differentiate both sides with respect to $t$
$$ \frac{d}{dt}\left(2i\sin t\right)^{2n} = 2i\sum_{k=0}^{2n}\binom{2n}{k}(-1)^k(n-k)e^{(2n-2k)it} $$
then isolate the contributes given by $k\in\{0,\ldots,n-1\}$ by multiplying both sides by $(e^{-2nit}+e^{-(2n-2)it}+\ldots+e^{-2it})$ (that is a simple geometric sum) and integrating over $(0,2\pi)$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\pi\,{\pars{-1}^{n + 1} \over 2^{2n - 3}}
\sum_{k = 0}^{n - 1}\pars{-1}^{k}{2n \choose k}\pars{n - k} =
\require{cancel}\cancel{{\pars{2n - 3}!! \over \pars{2n - 2}!!}\,{\pi \over 2}}:\ {\large ?}\,,
\qquad n \in \mathbb{N}_{\ >\ 0}}$.

$$
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\mbox{The RHS}\ correct\ answer\ \mbox{is}\quad 
{\pi \over 2^{2n - 3}}{2n - 2 \choose n - 1}}}
$$
\begin{align}
&\sum_{k = 0}^{n - 1}\pars{-1}^{k}{2n \choose k}\pars{n - k} =
\sum_{k = 0}^{n - 1}\pars{-1}^{k}\pars{n - k}{2n \choose 2n - k}
\\[5mm] = &\
\sum_{k = 0}^{n - 1}\pars{-1}^{k}\pars{n - k}\
\overbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z^{2n - k + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{2n \choose 2n - k}}
\\[5mm] = &\
\oint_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z^{2n + 1}}
\overbrace{\sum_{k = 0}^{n - 1}\pars{-1}^{k}\pars{n - k}z^{k}}
^{\ds{{n \over 1 + z} + {z \over \pars{1 + z}^{2}} + \pars{-1}^{n + 1}\,{z^{n + 1} \over \pars{1 + z}^{2}}}}\
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
n\
\underbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 1} \over z^{2n + 1}}
\,{\dd z \over 2\pi\ic}}_{\ds{{2n - 1 \choose 2n}\ =\ 0}}\ +\
\underbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 2} \over z^{2n}}
\,{\dd z \over 2\pi\ic}}_{\ds{{2n - 2 \choose 2n - 1}\ =\ 0}}\ +\
\pars{-1}^{n + 1}\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 2} \over z^{n}}
\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
\pars{-1}^{n + 1}{2n - 2 \choose n - 1}
\end{align}

$$
\bbx{\ds{\pi\,{\pars{-1}^{n + 1} \over 2^{2n - 3}}
\sum_{k = 0}^{n - 1}\pars{-1}^{k}{2n \choose k}\pars{n - k} =
{\pi \over 2^{2n - 3}}{2n - 2 \choose n - 1}}}
$$
