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
$$ f \in C^0(\mathbb R)\cap \left(\bigcap_{k=1}^\infty C^k(\mathbb R\setminus{\{0,1,\dots,k-1\}})\right).$$
I get:
$$f_1^{(1)}(x)=f_2(x)-f_1^2(x)$$
$$f_1^{(2)}(x)=2f_1^3(x)-3\,f_1(x)f_2(x)+f_3(x)$$
$$f_1^{(3)}(x)=-6f_1^4(x)+12f_1^2(x)f_2(x)-4f_1(x)f_3(x)-3f_2^2(x)+f_4(x)$$
$$f_1^{(4)}(x)=24f_1^5(x)-60f_1^3(x)f_2(x)+20f_1^2(x)f_3(x)+30f_1(x)f_2^2(x)-5f_1(x)f_4(x)-10f_2(x)f_3(x)+f_5(x)$$
$$f_1^{(5)}(x)=-120f_1^6(x)+360f_1^4(x)f_2(x)-120f_1^3(x)f_3(x)-270f_1^2(x)f_2^2(x)+30f_1^2(x)f_4(x)+120f_1(x)f_2(x)f_3(x)+30f_2^3
(x)-6f_1(x)f_5(x)-15f_2(x)f_4(x)-10f_3^2(x)+f_6(x)$$
$$...$$
Could you please correct and edit this in your question?
Clearly we could laborious try to derive a pattern first. But let's look first at The Online Encyclopedia of Integer Sequences (OEIS).
We get the integer sequence OEIS: A263634:
$$f_1^{(n)}(x)=\sum_{k=1}^{n+1}(-1)^{k-1}(k-1)!B_{n+1,k}(f_1(x),...,f_{n-k+2}(x)$$
$\ $
Now you can start proving this result.
To prove the result, you could use e.g. the answers at Simplify recurrence $\frac{d}{dx} f_{n-1}(x)= f_n(x)- f_{n-1}(x) f_1(x)$ :
Prove that the result are the logarithmic polynomials above. See Comtet, L.: Advanced Combinatorics. Reidel, 1974. p. 140: 3.5. Logarithmic and Potential Polynomials
Best Answer
The derivative of $x^{n}$ is $nx^{n-1}$, whatever integer is $n$. Thus the derivative of $(1-x)^n$ is $-n(1-x)^{n-1}$ and the sequence you get is therefore $$ \frac{1}{1-x}\quad \frac{1}{(1-x)^2}\quad \frac{2}{(1-x)^3}\quad \frac{6}{(1-x)^4}\quad \cdots $$ and a simple induction will confirm it: the $n$-th derivative is $$ f^{(n)}(x)=\frac{n!}{(1-x)^{n+1}} $$ Indeed, the derivative of the function $n!(1-x)^{-n-1}$ is $$ n!(-n-1)(-1)(1-x)^{-n-2}=\frac{(n+1)!}{(1-x)^{n+2}} $$