Simplifying algebraic sum with binomial coefficients

Question

I need to simplify this expression further:

$$ \sum_{m=1}^N (-1)^{m-1} m \binom{N}{m} $$

Answer 1

Hint: Recall that $$(1 - x)^N = \sum\limits_{m = 0}^N (-1)^m \binom{N}{m}x^m.$$

What happens if you differentiate both sides?

Answer 2

Use the identity $$ \binom{N}{m}\binom{m}{1} =\binom{N}{1}\binom{N-1}{m-1} =N\binom{N-1}{m-1} $$ to get that $$ \sum_{m=1}^N (-1)^{m-1} m \binom{N}{M} =\sum_{m=1}^N (-1)^{m-1}N\binom{N-1}{m-1} =N\sum_{u=0}^{N-1} (-1)^{u}\binom{N-1}{u}=0 $$ by considering the binomial expansion of $(1-1)^{N-1}$.