How to prove $\sum_{k=1}^n (-1)^{(k+1)} k \binom nk=0$ How to prove this identity?
$$\sum_{k=1}^n (-1)^{(k+1)} k \binom nk=0$$ This is George Casella statistical inference textbook exercise 1.27 (c). I have no idea to prove.
 A: \begin{align}
\sum_{k=1}^n (-1)^{k+1} k \binom{n}{k} 
&= n \sum_{k=1}^n (-1)^{k+1} \binom{n-1}{k-1} \\
&= n \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} \\
&= n (1-1)^{n-1} \\
&= n \cdot 0^{n-1} \\
&= \begin{cases}
1 & \text{if $n=1$} \\
0 & \text{otherwise}
\end{cases}
\end{align}
A: We assume $n > 1$.  Start with the Binomial Theorem:
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k$$
Differentiate:
$$n(1+x)^{n-1} = \sum_{k=1}^n k \binom{n}{k} x^{k-1}$$
Set $x=-1$ and observe that $(-1)^{k-1}=(-1)^{k+1}$:
$$\begin{align}
n(1-1)^{n-1} &= \sum_{k=1}^n k \binom{n}{k} (-1)^{k-1}\\
0 &= \sum_{k=1}^n k \binom{n}{k} (-1)^{k+1}
\end{align}$$
A: First we notice that the identity is not true for $n=1$. So we prove it for $n\geq\,2$. We use induction and the identity:
$\dbinom{n+1}{k}=\dbinom{n}{k}+\dbinom{n}{k-1}$.
Assume it is true for $n$. Now we write the $n+1$ statement as:
$\sum_{k=1}^{n+1}(-1)^{k+1}k\dbinom{n}{k}+\sum_{k=1}^{n+1}(-1)^{k+1}k\dbinom{n}{k-1}$.
Take the first term and notice that for $k=n+1$ the $n+1$ term is zero, by the definition of the factorial coefficient. Therefore we are left with
$\sum_{k=1}^{n}(-1)^{k+1}k\dbinom{n}{k}$ which is zero by the assumption of the induction.
Now we must prove that the second term is also zero. Set $k-1=m$ and get
$\sum_{m=0}^{n}(-1)^{m+2}(m+1)\dbinom{n}{m}$=$\sum_{m=0}^{n}(-1)^{m}(m+1)\dbinom{n}{m}$=$\sum_{m=0}^{n}(-1)^{m}m\dbinom{n}{m}+\sum_{m=0}^{n}(-1)^{m}\dbinom{n}{m}$ but $-\sum_{m=0}^{n}(-1)^{m+1}m\dbinom{n}{m}=0$
by the assumption of the induction (it makes no difference to add $m=0$).
So we are left with $\sum_{m=0}^{n}(-1)^{m}\dbinom{n}{m}$. But this is just (by Newton's binomial theorem) $[1+(-1)]^{n}=0$.
Thus the $n$-statement implies the $n+1$-statement and we are done!
