Let $f(x) = \tan ^{-1} x, x \in\mathbb R$, then the sequence $\{ f^{(n)} (0) \}$ is unbounded. True or false$?$ Let $f(x) = \tan ^{-1} x, x \in\mathbb R$, then the sequence
$\{ f^{(n)}(0) \}$ is unbounded. True or false$?$
Is there any short way to solve this?
Finding derivatives is really complicated after a few terms.

True is the answer,
and it gets short if the Taylor series
$$f(x)=\arctan x = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+1}$$
is available right from scratch. 
Then $f^{(n)}(0)\,$ can be read of from the Taylor coefficients: The non-zero values satisfy
$$\frac{f^{(2k+1)}(0)}{(2k+1)!} = \frac{(-1)^k}{2k+1}\,,$$
thus one has
$$f^{(n)}(0)\:=\, \begin{cases}0& \text{$n\,$ is even}\,,\\
(-1)^{(n-1)/2}\,(n-1)! & \text{$n\,$ is odd}\,.\end{cases}$$
