Show $\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}$ 
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*Given $$\sum\limits_{k=1}^n k{n\choose k} = n\cdot 2^{n-1}$$





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*I know that $$k\cdot{n\choose k}=n\cdot{n-1\choose k-1}=(n-k+1)\cdot{n\choose k-1}$$
Therefore $$\sum\limits_{k=1}^n k{n\choose k} = \sum\limits_{k=1}^n n{n-1\choose k-1} = n\cdot 2^{n-1}$$
So, $$n\cdot\sum\limits_{k=1}^n {n-1\choose k-1} = n\cdot 2^{n-1}$$
Therefore $$\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}$$

How is $\quad\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}\quad$?
 A: With $j=k-1$
$$\sum_{k=1}^n {n-1\choose k-1}=\sum_{j=0}^{n-1} {n-1\choose j}=2^{n-1} $$
A: $$2^{n-1}=(1+1)^{n-1}=\sum_{k=0}^{n-1}\binom{n-1}{k}\cdot 1^{n-1-k}\cdot 1^k=\sum_{k=0}^{n-1}\binom{n-1}{k}=\sum_{k=1}^{n}\binom{n-1}{k-1}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{k\ =\ 1}^{n}{n - 1 \choose k - 1} = 2^{n-1}:\ {\large ?}}$

With $\ds{a > 1}$:

\begin{align}
&\color{#66f}{\large\sum_{k\ =\ 1}^{n}{n - 1 \choose k - 1}}
=\sum_{k\ =\ 0}^{\infty}{n - 1 \choose k - 1}
=\sum_{k\ =\ 0}^{\infty}\bracks{%
\oint_{\verts{z}\ =\ a}{\pars{1 + z}^{n - 1} \over z^{k}}
\,{\dd z \over 2\pi\ic}}
\\[5mm]&=\oint_{\verts{z}\ =\ a}\pars{1 + z}^{n - 1}
\bracks{\sum_{k\ =\ 0}^{\infty}\pars{1 \over z}^{k}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a}\pars{1 + z}^{n - 1}
\bracks{{1 \over 1 - 1/z}}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ a}{z\pars{1 + z}^{n - 1} \over z - 1}
\,{\dd z \over 2\pi\ic}
=2\pi\ic\lim_{z\ \to\ 1}\bracks{\pars{z - 1}{z\pars{1 + z}^{n - 1} \over z - 1}
\,{1 \over 2\pi\ic}}
=\color{#66f}{\LARGE 2^{n - 1}}
\end{align}
A: Combinatorially we can prove this simply by noting that both the lefthand side and the righthand side count subsets of $\{1,2,\dots,n\}$ all containing a fixed element, $1$. 
