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 $\quad\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}\quad$?

With $j=k-1$
$$\sum_{k=1}^n {n-1\choose k-1}=\sum_{j=0}^{n-1} {n-1\choose j}=2^{n-1} $$
