Suppose $f$ is entire, $\lambda$ is a positive number, and the inequality $|f(z)|\lt \exp(|z|^{\lambda})$ holds for all large enough $|z|$. If $f(z) = \sum a_nz^n$, prove that the inequality $|a_n|\leq \Bigl(\frac{e\lambda}{n}\Bigl)^{n/\lambda}$ holds for all large enough $n$. (Problem $15.2$ from Rudin's RCA; not a homework, I'm just doing it for fun.)
Let $r$ be a sufficiently large positive number and $\gamma(t)=re^{it}, 0\leq t\leq 2\pi.$ Then, $$|a_n|=|f^{(n)}(0)|=\left|\frac{1}{2πi}\int_{\gamma} \frac{f(z)}{z^{n+1}}dz\right|\leq \frac{\exp(r^{\lambda})}{r^n}.$$ So for large enough $n$, I have $$|a_n|\leq \frac{\exp(n^{\lambda})}{n^n}.$$ I don't know what to do next. One another thing I noticed is that if I take $g(r)=\frac{\exp(r^{\lambda})}{r^n}$ and solve $g{'}(r_0)=0$, I get $r_0=\left(\frac{n}{\lambda}\right)^{1/\lambda}$ and $g(r_0)=\Bigl(\frac{e\lambda}{n}\Bigl)^{n/\lambda}.$ But, that's just some calculus stuff and I think it's not rigorous enough. So how do I solve this with rigour?