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Is there $f$ holomorphic on unit disc, such that $f^{(n)}(0)\ge n^{2n}$ for all $n\ge 0$? (Is it possible that it was intended that $|f^{(n)}(0)|\ge n^{2n}$?)

What I have done so far is: Look at the Laurent series of $f(z)$ about $z=0$. For $n\le -1$, $a_n={1\over 2\pi i}\int_{|z|=1}{f(z)\over z^n}dz=0$ by Cauchy's Theorem. So $a_n={f^{(n)}(0)\over n!}$ which makes the Laurent Series a Taylor series. Radius of convergence: ${1\over R}=\limsup_{n\to \infty}|a_n|^{1\over n}\ge \limsup_{n\to \infty}{n^2\over n!^{1\over n}}\ge \limsup_{n\to \infty}{n^2\over n}=\lim_{n\to \infty}{n}=\infty$. I am really not confident in what I have done. Am I allowed to conclude that $R=0$? What should be the final conclusion?

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    Sure. And $R = 0$ contradics the fact that $f$ is holomorphic on the unit disc, which should imply $R \geq 1$. – 2017-01-08
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    So what I did is legitimate? Does it all come down to the radius of convergence? – 2017-01-08
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    Yes, your approach looks fine in view of the Cauchy-Hadamard theorem. – 2017-01-08
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    Thank you for helping me sorting it out. – 2017-01-08
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    Also, in order to get to a contradiction, you may want to use the fact that $$R = \sup \{ r > 0 : f(z) \text{ is holomorphic on } |z| < r \}. $$ Together with the assumption, this gives $R \geq 1$ which contradicts $R= 0$. – 2017-01-08

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