Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$? Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$ ?

I know that $\sum_{k=0}^{n}\binom{n}{k}= 2^{n}$ and so, $\sum_{k=1}^{n}\binom{n}{k}= 2^{n}-1$ but how to deal with $k$ ?
 A: From the binomial theorem, we have
$$(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k\tag 1$$
Differentiating $(1)$ reveals
$$n(1+x)^{n-1}=\sum_{k=0}^n\binom{n}{k}kx^{k-1}\tag2$$
Setting $x=1$ in $(2)$ yields 
$$n2^{n-1}=\sum_{k=0}^n\binom{n}{k}k$$
And we are done!

Interestingly, I showed in THIS ANSWER, that for $m<n$, we have $$\sum_{k=0}^n\binom{n}{k}(-1)^k k^m=0$$
A: There is also a combinatorial argument:
Suppose you have a room of $n$ people and want to select a committee of $k$ of them, where one member is the chairperson. There are $k\binom{n}{k}$ ways to do this. Your sum represents the total number of ways to select such a committee of any size (from $1$ to $n$) with a chairperson.
How else can we think of this? Instead, first pick the committee chairperson. There are $n$ ways to do this. Then, go to each of the remaining $n-1$ people and decide if they should be in the committee. There are $2^{n-1}$ ways to do this. Note that we can create any committee/chairperson team this way, as before. Hence your sum is equal to $n 2^{n-1}$.
A: We have
\begin{align*}
\sum_{k=0}^nk{n\choose k}
&=\sum_{k=1}^nk{n\choose k}\\
&=n\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}\\
&=n\sum_{\ell=0}^{n-1}\frac{(n-1)!}{\ell!((n-1)-\ell)!}
\tag{by taking $\ell=k-1$}\\
&=n\sum_{\ell=0}^{n-1}{n-1\choose\ell}\\
&=n2^{n-1}.
\end{align*}
A: Proof without derivatives:
$$\sum_{k=1}^nk\binom{n}{k}=\sum_{k=1}^n k\frac{n!}{(n-k)!k!} =\sum_{k=1}^n \frac{n!}{(n-k)!(k-1)!} \\
= \sum_{k=1}^n \frac{n(n-1)!}{(n-k)!(k-1)!} = \sum_{k=1}^n n\binom{n-1}{k-1} \\
= n \sum_{k=0}^{n-1}\binom{n-1}{k}= n2^{n-1}$$
Alternate proof via probability theory:
Toss a fair coin $n$ times, find the expected no of heads. Let $N$ be the random variable denoting the number of heads.
Then $E[N] = n/2$ because $N$ is the sum of $n$ bernoulli random variables with probability $1/2$. But we also know that $N$ has a binomial distribution.
Hence
$$E[N] =\sum_{k=1}^nk\binom{n}{k}2^{-n} $$
Rearrange to get your answer.
A: "And so $\sum_{k=1}^n\binom nk=2^{n-1}$". Nope: $2^n-1$.
Sketch: Notice that $$\sum_{k=1}^n k\binom nk x^{k-1}$$ is the derivative of $\sum_{k=0}^n\binom nk x^k$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\mc{I} & \equiv \sum_{k = 1}^{n}k{n \choose k} =
\sum_{k = 0}^{n}\pars{n - k}{n \choose n - k} =
n\sum_{k = 0}^{n}{n \choose k} - \sum_{k = 0}^{n}k{n \choose k} =
n\ 2^{n} - \sum_{k = 1}^{n}k{n \choose k}
\\[5mm] & = 2^{n}\,n - \mc{I} \implies
\bbx{\ds{\mc{I} \equiv \sum_{k = 1}^{n}k{n \choose k} = 2^{n - 1}\,n}}
\end{align}
