Proving $f^{(n)} (0)$ s unbounded where $f(x)= \tan^{-1} x$

Question

The following question was asked in real analysis quiz and I am looking for an more elegant solution.

Question :Let $f(x)= \tan^{-1}x $ for all $x\in \mathbb{R}$. Prove that sequence $\{f^{(n)}(0)\}$ is unbounded.

There is a solution given by differentiating $n$ times the expression $(1+x^2) f(x)$ and then using the recurrence relation $(1+x^2) f^{(n+1)} (x)+ 2nx f^{(n)} (x) + n(n-1) f^{(n-1)} (x)=0$ and $f'(0)=1$. But is there a more elegant solution using any other concepts of other branches of pure mathematics or analysis (real, complex, functional).

I thought I should ask here as I don't wanted to use that method although that is clear to me!

Answer 1

From $$f'(x)={1\over 1+x^2}=1-x^2+x^4-x^6+x^8-\ldots\qquad(-1<x<1)$$ we obtain $$f^{(2n+1)}(0)=(-1)^n(2n)!\quad .$$

Answer 2

$$f'(x)=\frac{1}{1+x^2}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$$ Use $$D^n(x+a)^{-1}=(-1)^n n! (x+a)^{-n-1}$$ Then $$f^{n}(x)=(2i)^{-1} (-1)^{n-1} (n-1)! \left(\frac{1}{(x+i)^{n}}-\frac{1}{(x-i)^{n}}\right).$$ $$f^{n}(0)=\frac{(-1)^n-1}{2i^{n+1}} (n-1)!$$