Any other ways to evaluate $\displaystyle \sum_{k=0}^n k {n \choose k} p^k (1-p)^{n-k}$? 
$\displaystyle \sum_{k=0}^n k {n \choose k} p^k (1-p)^{n-k}$ with $0<p<1$

I know of one way to evaluate it (from statistics) but I was wondering if there are any other ways. 
This is the way I know:
Let 
$$M(t)=\displaystyle \sum_{k=0}^n e^{kt} {n \choose k} p^k (1-p)^{n-k}$$
Then $$M(t)=\displaystyle \sum_{k=0}^n {n \choose k} (pe^t)^k (1-p)^{n-k}=(pe^t+1-p)^n$$
$$M'(t)=\displaystyle \sum_{k=0}^n ke^{kt} {n \choose k} p^k (1-p)^{n-k}=pe^tn(pe^t+1-p)^{n-1}$$
$$M'(0)=\displaystyle \sum_{k=0}^n k {n \choose k} p^k (1-p)^{n-k}=np$$
 A: Yes.
$$\begin{align}
\sum_{k=0}^n k \binom{n}{k} p^k (1-p)^{n-k}
&= \sum_{k=1}^n k \binom{n}{k} p^k (1-p)^{n-k}
= \sum_{k=1}^n k\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}\\
&=\sum_{k=1}^n \frac{n!}{(k-1)!(n-1-(k-1))!} p^k (1-p)^{n-k}\\
&=np\sum_{k=1}^n \frac{(n-1)!}{(k-1)!(n-1-(k-1))!} p^{k-1} (1-p)^{n-1-(k-1)}\\
&=np\sum_{\ell=0}^{n-1} \frac{(n-1)!}{\ell!(n-1-\ell)!} p^{\ell} (1-p)^{n-1-\ell}\\
&=np\sum_{\ell=0}^{n-1} \binom{n-1}{\ell} p^{\ell} (1-p)^{n-1-\ell}\\
&= np.
\end{align}$$
A: A slightly different variation of an answer already given which might also be convenient.

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k}}
&=np(1-p)^{n-1}\sum_{k=1}^n\binom{n-1}{k-1}\left(\frac{p}{1-p}\right)^{k-1}\tag{1}\\
&=np(1-p)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}{k}\left(\frac{p}{1-p}\right)^{k}\tag{2}\\
&=np(1-p)^{n-1}\left(1+\frac{p}{1-p}\right)^{n-1}\tag{3}\\
&\color{blue}{=np}\tag{4}
\end{align*}

Comment:


*

*In (1) we use the binomial identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and we factor out $p(1-p)^{n-1}$ to prepare the index shift of the next line.

*In (2) we shift the index to start from $k=0$.

*In (3) we use the binomial summation formula.

*In (4) we do some final simplifications.
A: You can also notice that this is the expectation $E(X)$ where $X$ is a binomial random variable with parameters $n$ and $p$, and $E(X)=n\,p$.
