Calculate $f^\prime(0)$ of $f(x) = \prod_{n=0}^{100} (x-n)$. How can one calculate $f^\prime(0)$ of $f(x) = \prod_{n=0}^{100} (x-n)$ by hand?
I tried to compute the derivative of $x$ and got $1$, and the derivative of $x(x-1)$ at $0$ is $-1$, and the derivative of $x(x-1)(x-2)$ at $0$ is $2$, and the derivative of $x(x-1)(x-2)(x-3)$ is $-6$. I don't see a pattern, and I'm not sure of an alternative route. 
 A: Using definition
$$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{\prod_{n=0}^{100}(x-n)}{x}=\prod_{n=1}^{100}(-n)=100!.$$
A: Hint: the function $f(x)$ is a polynomial of order $101$. When you differentiate it and then set $x=0$ the only term left is the coefficient in front of $x$ in $f$. Try to figure out how the term with $x$ is build here.
A: By writting $$f(x)=xg(x),\qquad\text{where}\;g(x)=\prod_{n=1}^{100}(x-n)$$
we get
\begin{align*}
f'(x)&=g(x)+xg'(x)\qquad \implies \qquad f'(0)=g(0)+0=\prod_{n=1}^{100}(-n)=100!
\end{align*}
A: $$\ln  f(x)=\sum_{n=0}^{100}\ln(x-n)$$
$$\implies f'(x)=f(x)\sum_{n=0}^{100}\dfrac1{x-n}=\prod_{n=1}^{100}(x-n)+\text{terms containing }x$$
$$\implies f'(0)=?$$
A: $f(x)$ is a polynomial with degree $100$ in the $x$ variable, hence $f'(0)$ is just the coefficient $[x^1]\,f(x)$, that also is
$$ [x^0]\frac{f(x)}{x}=[x^0]\prod_{n=1}^{100}(x-n) = \left.\prod_{n=1}^{100}(x-n)\right|_{x=0} = \prod_{n=1}^{100}(-n) = \color{red}{100!}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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There is a closed expression for $\ds{\,\mrm{f}\pars{x}}$ which can be straightforward derived:

\begin{align}
\mrm{f}\pars{x} & \equiv \prod_{n = 0}^{100}\pars{x - n} =
-\prod_{n = 0}^{100}\pars{n - x} = -\pars{-x}^{\overline{101}} =
-\,{\Gamma\pars{-x + 101} \over \Gamma\pars{-x}}
\\[5mm] & =
\bbx{\ds{{\sin\pars{\pi x} \over \pi}\,\Gamma\pars{101 - x}\Gamma\pars{1 + x} =
\pars{\color{#f00}{100!}\,x}\,{\sin\pars{\pi x} \over \pi x}
\,{1 \over {100 \choose x}}}}
\end{align}

Obviously, the linear term is $\bbx{\ds{100!}}$.

