Roots of $f$ and $f'$ for $1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$ I have this question from an admission exams. Given that $$f(x)=1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively $f'(x)$
Here is how I tried, since for $x=1,2,\ldots,100, 101$ the sum part vanishes little by little and
$$f(1)=1-1=0$$
$$f(2)=1-2+\frac{1}{2!}2(2-1)=0$$
$$f(3)=1-3+\frac{1}{2!}3(3-1)-\frac{1}{3!}3(3-2)(3-1)=\binom{3}{0}-\binom{3}{1}+\binom{3}{2}-\binom{3}{3}$$ and we can see a pattern in the above equation so that we can rewrite $f(3)$ as $(1-1)^3=0\,$ and indeed that $f(101)=(1-1)^{101}=0$ 
Since the polynomial is also of order $101$, the roots are $x=1,2, \ldots, 101$ giving:$$S(f(x))=\sum_{j=1}^{101}j =5151$$ Is this correct? And how can  $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $$f(x)=-\frac{1}{101!}(x-1)(x-2)\cdots(x-101)$$ 

We have
$$f(x) = a_{101}x^{101} + a_{100}x^{100} + \cdots$$
and we know that $a_{101} = -\frac1{101!}$. Vieta's formulas give
 $$5151 = \text{ sum of roots of } f =  -\frac{a_{100}}{a_{101}}$$
so $a_{100} = \frac{5151}{101!}$.
The derivative is
$$f'(x) = 101a_{101}x^{100} + 100a_{100}x^{99} + \cdots$$
so Vieta's formulas give
 $$\text{ sum of roots of } f' =  -\frac{100a_{100}}{101a_{101}} = -\frac{100\cdot \frac{5151}{101!}}{101\cdot \frac{-1}{101!}} = \frac{515100}{101}= 5100$$
