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 $$f(x)=-\frac{1}{101!}(x-1)(x-2)\cdots(x-101)$$ 

According to Vieta given the polynomial
$$
p_n(x) = a_n x^n+a_{n-1}x^{n-1}+\cdots + a_0\\
p'_n(x) = n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2}+\cdots+a_1
$$
now 
$$
S_{100}(x) = -\frac{a_{100}}{a_{101}}\\
S'_{100}(x) = -\frac{(101-1)a_{100}}{101 a_{101}}
$$
Here
$$
a_{101} = \frac{(-1)^{101}}{101!}\\
a_{100} = -a_{101}\frac{100(100+1)}{2}+\frac{(-1)^{100}}{100!}
$$
then
$$
S_n(x) = 5151\\
S'_n(x) = 5100
$$
and finally
$$
S_{100}(x)-S'_{100}(x) = 5151-5100 = 51
$$
