Further explanation for steps of an equation that proofs that $\sum^{n}_{k=0}k\cdot \binom{n}{k}=n\cdot2^{n-1}$ So, this is one of the questions in my textbook, which seems to be quite common: $$\sum^{n}_{k=0}k\cdot \binom{n}{k}=n\cdot2^{n-1}$$
The same book provides the following solution: 
$$\sum_{k=0}^{n}k\cdot \binom{n}{k}= \sum^{n}_{k=0}n\cdot\binom{n-1}{k-1}=n\sum^{n-1}_{k=0}\binom{n-1}{k}=n\cdot2^{n-1}$$
What is unclear to me, is (1) how I get from $$\sum^{n}_{k=0}n\cdot\binom{n-1}{k-1} $$ to$$n\sum^{n-1}_{k=0}\binom{n-1}{k}$$ and (2) from $$n\sum^{n-1}_{k=0}\binom{n-1}{k}$$ to $$n\cdot2^{n-1}$$ respectively.
 A: (1) We have $$\sum_{k=0}^nn\cdot{n-1\choose k-1}=n\sum_{k=0}^{n-1}{n-1\choose k}$$ by a substitution $k'=k-1$. We can obviously place the $n$ at the front, and we then use ${n-1\choose-1}=0$, so $k=0$ can be ignored.
(2) Here we just use the formula $\sum_{k=0}^n{n\choose k}=2^n$. There is a very nice combinatorial proof of this. The term ${n\choose k}$ represents the amount of ways to pick $k$ items out of $n$. If we sum over all $0\leq k\leq n$, we just look at the total amount of ways to pick any set out of $n$ items. There are $2^n$ ways, since we can choose for each item whether we pick it or not.
There is also a very nice combinatorial proof of the statement you want to prove in the first place. The term $k\cdot{n\choose k}$ can be thought of the amount of ways to pick $k$ items out of $n$ and then choosing one of the $k$ items as your favorite. Summing over $0\leq k\leq n$, we just look at the total amount of ways to pick any set out of $n$ items, and then choosing one of the picked items as your favorite. We can also reverse this: First choose your favorite item, and then pick any set of remaining items out of the $n-1$ left over items. This results in $n\cdot2^{n-1}$ possibilities.
A: The second part is very simple. As the $\sum^{n-1}_{k=0}\binom{n-1}{k}$ is the size of power set of a set with size of $n-1$. Hence, it is equal to $2^{n-1}$. 
Hint: For the first part, there is a mistake in the range of $\sum$. If it would be true, you can show it using a simple change variable ($u = k - 1$ and change the range of sigma from ($1$ to $n$) to ($0$ to $n-1$)).
