Binomial coefficients-sums I need help solving this task so if anyone had a similar problem it would help me.
Calculate:
$\sum\limits_{i=0}^{n}(-1)^i i {n \choose i}$
I tried this:
$\sum\limits_{i=0}^{n}(-1)^i i \frac{n!}{i!(n-i)!}\\\sum\limits_{i=0}^{n}(-1)^i \frac{n!}{(i-1)!(n-i)!}\\\sum\limits_{i=0}^{n}\frac{(-1)^i n!}{(i-1)!(n-i)!}$
And now with this part I don’t know what to do next.
Thanks in advance !

I did it like this, so I'm looking for your thoughts, is it right?
$\sum\limits_{i=0}^{n} (-1)^{i} i\binom {n} {i}\\\sum\limits_{i=0}^{n} (-1)^{i} n\binom {n-1} {i-1}\\\sum\limits_{i=0}^{n} (-1)^{i} n\frac{(n-1)!}{(i-1)!(n-i)!}    \\\sum\limits_{i=0}^{n} (-1)^{i} n(n-1) \frac{1}{(i-1)!(n-i)!}\\n(n-1)!\sum\limits_{i=0}^{n} (-1)^{i}\frac{1}{(i-1)!(n-i)!}\\ 
n\sum\limits_{i=0}^{n} (-1)^{i}\frac{(n-1)!}{(i-1)!(n-i)!}\\n\sum\limits_{i=0}^{n}\binom{n-1}{i-1}(-1)^{i}\\n(1-x)^{n-1}=0 $ $n>1,n>-1,n=1$
Thanks for help !
