Derivatives of Infinite Product that Diverges to 0 Consider the function given by $$f_n(x) = \prod\limits_{k=2}^n \left( 1-\frac{x}{k} \right).$$ Then for all $x \in (0,2)$, we have that $$\lim_{n\to\infty}\; f_n(x) = \prod\limits_{k=2}^\infty \left( 1-\frac{x}{k} \right) =0. $$
This follows from the fact that if $q_k \in [0,1)$ for all $k$, then $\prod\limits_{k=1}^{\infty} (1-q_k) = 0$ if and only if $\sum\limits_{k=1}^\infty q_k$ diverges (See for example here and here).
I am interested in the derivatives of this function $f_n$. For instance, let $f_n^{(j)}$ denote the $j$-th derivative of $f_n$. Is it true that $\lim_{n\to\infty} f_n^{(j)}(x) = 0$ for all $x\in(0,2)$? 
Considering the first derivative, we have that from the product rule:
$$ f_n'(x) =  \sum\limits_{k=2}^n \frac{-1}{k} \prod\limits_{i \neq k} \left(1-\frac{x}{i} \right)\tag1 \label 1 .$$
Now, we have that 
\begin{align}\prod\limits_{i \neq k} \left(1-\frac{x}{i} \right) &\leq \prod\limits_{i =2}^{n-1} \left(1-\frac{x}{i} \right)\\
&= \exp\left\{ \sum_{i=2}^{n-1} \log\left(1-\frac{x}{i} \right)  \right\} \\
&\leq \exp\left\{ \sum_{i=2}^{n-1} -\frac{x}{i} \right\} \tag2  \\
&\leq \exp\left\{-x\log(n)+x \right\}\tag3\\
&= {\left(\frac{e}{n}\right)}^x.\end{align}
Where (2) above follows from the fact that $\log(1-y)\leq -y$ for all $y<1$, and (3) follows from $\sum\limits_{i=2}^{n-1}\frac{1}{i} \geq \log(n)-1$. Now, by noticing all the terms in the sum in $\ref1$ are nonpositive, we have the following
\begin{align*}
 f_n'(x) \geq {\left(\frac{e}{n}\right)}^x\sum\limits_{k=2}^n \frac{-1}{k} \geq -{\left(\frac{e}{n}\right)}^x \log(n)
\end{align*}
Finally, because $f_n'(x)\leq 0$ and $\lim_{n \to \infty}  {\left(\frac{e}{n}\right)}^x \log(n) = 0$, we are able to conclude that $\lim \; f_n'(x) = 0$ for all $x \in (0,2)$. 
Now, is there a way to show this for the $j$-th derivative $f_n^{(j)}$? I am unable to even currently show that the second derivative is zero (in my simulations it seems to approach zero, but at a rate slower than the first derivative). An idea that I have briefly considered using is considering the differentiation of the infinite product (see here and similarly here). However in my case my infinite product diverges to zero, so I am not sure how useful this will be. Also, I am interested in the higher order derivatives of this infinite product, not just the first derivative. 
I appreciate any input on this problem! 
 A: "The shortest path to truth in $\Bbb R$ is through $\Bbb C.$" -- Hadamard. 
We can use a powerful but basic result from complex analysis to quickly answer this: Extend the formula for each $f_n$ to the domain  $D=\{z\in \Bbb C: |z-1|<1\}\cup \{z\in \Bbb C: |z-2|<1\}.$ 
Then each $f_n$ is analytic on $D$ and the sequence $(f_n)_{n\geq 2}$ converges uniformly to $0$ on any $E\subset D$ such that $E$ is a closed subset of $\Bbb C.$   For  $x\in D$  take $r_x>0$ such that $\{z: |z-x|\leq r_x\}\subset D.$ 
For any $j\geq 0$ we have  $$f_n^{(j)}(x)=\frac {j!}{2\pi i}\int_{|z-x|=r_x}\frac {f_n(z)}{(z-x)^{j+1}}dz.$$ For fixed $j$ we have  $|f_n(z)|\to 0$ uniformly on $\{z:|z-x|=r_x\}$ as $n\to \infty,$   so $f_n^{(j)}(x)\to  0$ as $n\to \infty.$ 
Moreover, as $n\to\infty,$  for each $j\geq 0$ the sequence  $(\;f_n^{(j)}\;)_{n\geq 2}$ converges uniformly to $0$ on  $[s,2]$ for any $s>0.$
This may be outside the scope of your background but it shows the  strength of  complex analysis applied to real analysis.  
A: as a HINT, notice that
$$
\eqalign{
  & f(x,n) = \prod\limits_{2\, \le \,k\, \le \;n} {\left( {1 - {x \over k}} \right)} 
  = {{\left( { - 1} \right)^{\,n - 1} \prod\limits_{2\, \le \,k\, \le \;n} {\left( {x - k} \right)} } \over {\prod\limits_{2\, \le \,k\, \le \;n} k }} =   \cr 
  &  = {{\left( { - 1} \right)^{\,n - 1} \prod\limits_{0\, \le \,k\, \le \;n - 2} {\left( {x - 2 - k} \right)} } \over {n!}} 
  = {{\left( { - 1} \right)^{\,n - 1} \left( {x - 2} \right)^{\,\underline {\,n - 1\,} } } \over {n!}} =   \cr 
  &  = {{\left( { - 1} \right)^{\,n - 1} } \over {x - 1}}\left( \matrix{  x - 1 \cr  n \cr}  \right)
  = \left( { - 1} \right)^{\,n - 1} {{\Gamma \left( {x - 1} \right)} \over {\Gamma \left( {x - n} \right)\Gamma \left( {n + 1} \right)}} \cr} 
$$
and also
$$
\eqalign{
  & f(x,n) = {{\left( { - 1} \right)^{\,n - 1} } \over {x - 1}}\left( \matrix{  x - 1 \cr   n \cr}  \right)
 = {1 \over {1 - x}}\left( \matrix{  n - x \cr   n \cr}  \right)
 = {1 \over {1 - x}}\left( \matrix{  n - x \cr    - x \cr}  \right) =   \cr 
  &  = {1 \over {1 - x}}{{\Gamma (n - x + 1)} \over {\Gamma (n + 1)\Gamma (1 - x)}} = {{\Gamma (n + 1 - x)} \over {\Gamma (n + 1)\Gamma (2 - x)}} \cr} 
$$
