Help understanding the weak law of large numbers with respect to statistics I'm trying to do some self-studying to upgrade my statistics knowledge, and came across this term in a section discussing the weak law of large numbers and Bernoulli's theorem:
$$\sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$$
According to the book I was reading, this term can "easily be shown to equal np". 
I am at a loss as to how to do so and could use some guidance!
Edited
Lower limit $k=0$, not $1$
 A: Let $X_n=B(n,p)$ be a binomially distributed random variable. Also notice that $X_n=Y_1+Y_2+\cdots+ Y_n$ where $Y_i$ are i.i.d. Bernoulli with parameter $p$.
Now observe that
\begin{align}
\sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}&= \operatorname{E}(X_n)\\
&= \operatorname{E}(  Y_1+Y_2+\cdots Y_n)\\
&=\operatorname{E}(  Y_1)+\operatorname{E}(Y_2)+\cdots +\operatorname{E}(Y_n)\\
&=np
\end{align}
A: We have this sum:
$$
\sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \tag 1
$$
First notice that when $k=0$, the term $k\dfrac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$ is $0$, and next notice that when $k\ne0$ then
$$
\frac k {k!} = \frac 1 {(k-1)!}
$$
so that
$$
k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} = \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}.
$$
The two expressions inside the parentheses in the denominator now add up to $n-1$ rather $\text{than } n.$  So we can write it like this:
$$
\frac{n!}{(k-1)!(n-k)!} = n\cdot \frac{(n-1)!}{(k-1)!(n-k)!} = n\cdot \binom{n-1}{k-1}.
$$
Then the sum $(1)$ becomes
$$
\sum_{k=1}^n n\cdot \binom{n-1}{k-1} p^k (1-p)^{n-k}.
$$
Since $n$ does not change as $k$ goes from $1$ to $n$, we can pull $n$ out, getting
$$
n \sum_{k=1}^n \binom{n-1}{k-1} p^k (1-p)^{n-k}.
$$
Now let $j=k-1$ and observe that as $k$ goes from $1$ to $n$ then $j$ goes from $0$ to $n-1$, and $k = j+1$, so we have
$$
n \sum_{j=0}^{n-1} \binom{n-1} j  p^{j+1} (1-p)^{(n-1)-j}.
$$
Since $p$ does not change as $j$ goes from $0$ to $n-1$, we can pull $p$ out, getting
$$
np \sum_{j=0}^{n-1} \binom{n-1} j  p^j (1-p)^{(n-1)-j}.
$$
Now let $m= n-1$, so we have
$$
np \sum_{j=0}^m \binom m j p^j (1-p)^{m-j}.
$$
This sum is $1$ since it's the sum of probabilities assigned by the $\mathrm{Binomial}(m,p)$ distribution.  Hence we get
$$
np\cdot 1.
$$
A: The trick is the using the identity $k { n \choose k} = n {n-1 \choose k-1}$. 
$$\begin{align*}
&\sum_{k=1}^n k { n \choose k } p^k (1-p)^{n-k}\\
&= \sum_{k=1}^n n { n-1 \choose k-1} p^k (1-p)^{n-k}\\
&=np \sum_{k=1}^n {n-1 \choose k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\
&=np (p+(1-p))^n\\
&=np
\end{align*}$$
