How to prove this result on binomial coefficients?

Solution 1:

$$\sum_{j \mathop = 1}^n \left({-1}\right)^{n + 1} j \binom n j=\sum_{j \mathop = 1}^n \left({-1}\right)^{n + 1} n \binom {n - 1} {j - 1}= n \sum_{j \mathop = 0}^{n - 1} \left({-1}\right)^{n - 1} \binom {n - 1} j=0$$

NOTE

we have used that

$$j \binom n j= n \binom {n - 1} {j - 1}$$

which has a simple explanation: select $j$ people out of n, then designate one as special. The LHS represents how many ways we can do this by first picking the $j$ people and then making designation. On the RHS, we have the number of ways to select the special one and then picking the remaining $j-1$ from the remaining $n-1$.