Construct a function with each derivative being non-differentiable at a distinct point 
Construct a function $f_1:\mathbb{R}\to\mathbb{R}$ with the following properties or show that no such function exists:
$1.$ $f_1$ is differentiable everywhere except one point $x_1.$
$2.$ Define $f_2 : \mathbb{R}\setminus\{x_1\} \to \mathbb{R}$ as $f_2(x) := $ derivative of $f_1$ at $x.$ This $f_2$ must be differentiable everywhere in its domain except one point $x_2.$
$3.$ Define $f_3 : \mathbb{R}\setminus\{x_1,\;x_2\} \to \mathbb{R}$ as $f_3(x) := $ derivative of $f_2$ at $x.$ This $f_3$ must be differentiable everywhere in its domain except one point $x_3.$
$\vdots$
$n.$ Define $f_n : \mathbb{R}\setminus\{x_1, \cdots, x_{n-1}\} \to \mathbb{R}$ as $f_n(x) := $ derivative of $f_{n-1}$ at $x.$ This $f_n$ must be differentiable everywhere in its domain except one point $x_n.$
$\vdots$
(Note that we do not stop at any $n.$) 

I found this question in a collection of extra questions for my Calculus course.
I started out by trying something along the lines of $f(x) = \lim_{n\to\infty}\sum_{i=1}^n (x-i)^i|x-i|$, but the function itself isn't defined anywhere and I couldn't figure out how to fix it with minimum effort.
So next, I tried out something that might actually be defined somewhere such as $$f(x) = \lim_{n\to\infty}\sum_{i=1}^n \frac{\left(\frac{2}{\pi}\arctan(x-i)\right)^i\left(\frac{2}{\pi}\arctan|x-i|\right)}{(i+1)!}$$
which is defined for $x \in \mathbb{R}$, but I wasn't able to prove continuity or differentiability. Intuitively, I feel that since it's a sum of continuous functions, it should be continuous, but I'm not sure whether this intuition is correct because it's an infinite sum.
I asked the person whose website I found the question on (another student), and he said that he wasn't sure whether such a function was even possible.
Any help would be appreciated!
 A: A different solution is the following 'lazy exponential' - there are easier solutions (maybe look up bump functions), but I like delay ODEs. set
\begin{align}x\in(-\infty,0]&\implies f(x):=1,\\
x\in(0,1] &\implies f(x) := 1+x, \\
x\in (1,2] &\implies f(x) := 1+x + \frac{(x-1)^2}{2!},\\
x\in(2,3] & \implies f(x) := 1+x + \frac{(x-1)^2}{2!} + \frac{(x-2)^3}{3!},
\end{align}
and in general
$$x\in(n,n+1]\implies f(x) := \sum_{k=0}^{n+1} \frac{(x-k+1)^k}{k!}. $$
If you differentiate, you find for $x\in (n,n+1)$, where $n>1$:
$$ f'(x) = \sum_{k=1}^{n+1} \frac{(x-k+1)^{k-1}}{(k-1)!} =\sum_{j=0}^{n} \frac{(x-j)^{j}}{j!}= f(x-1)$$
so to the right of 1, it solves a delay ODE with initial data prescribed on $x\in(0,1]$ above. $f'$ is clearly discontinuous at $0$, but $$\left.\frac{d}{dx}\frac{(x-1)^2}{2!}\right|_{x=1} = 0 $$ so the derivative is continuous at $x=1$. In general, for any integer $n\ge 2$, near $x=n-1$, all the terms  $\frac{(x-h+1)^h}{h!}$ for $h<n$ are smooth, and the newly added term $T_n$,
$$ T_n(x) := \begin{cases} \frac{(x-n+1)^n}{n!} & x>n-1,\\ 0 & x\le n-1\end{cases}$$ 
is $C^1$. Conclusion - $$f \in C^0(\mathbb R)\cap C^1(\mathbb R\setminus \{0\}).$$
To finish, we use the delay ODE, which says that differentiating is the same as translating the function to the right by one. Thus
 for $x\in \mathbb (0,\infty)\setminus \mathbb N$, $i\in\mathbb N$,
$$ f^{(i+1)}(x+i) = f'(x).$$
So the discontinuity of $f^{(i+1)}$ at $x=i-1$, and the continuity at integers $x=\tilde i > i-1$ follows directly from the dis/continuity of $f'$ at $0,1,2,\dots$. We conclude
$$ f \in C^0(\mathbb R)\cap \left(\bigcap_{k=1}^\infty C^k(\mathbb R\setminus{\{0,1,\dots,k-1\}})\right).$$
