How to prove that, if $|f(z)| \le C \log^{p} (|z| + 1) $ for some $C \ge 0$ and $p>0$, $f(z) = 0$ for all $z \in \mathbb{C}$? Let $f: \mathbb{C} \to \mathbb{C} $ be analytic and assume that, for some $C \ge 0$ and $p > 0 $, $|f(z)| \le C \log^{p} (|z| + 1)  $ for all $z \in \mathbb{C} $, in which $\log: \mathbb{R}_{>0} \to \mathbb{R} $ is the usual logarithm. How does one show that $f$ is the zero-function?
This is what I tried: As per Cauchy's Power Series expansion theorem (1831), we know that if $f$ can be represented as some sort of formal power series: $$ f(z) = \sum_{n=0}^{\infty} a_n (z-a)^n ,$$ in which the coefficients $a_n$ can be represented as follows: $$ |a_n| = \left| \frac{1}{2 \pi i} \oint_{\left| \zeta - a \right| = \rho } \frac{f(\zeta) }{ (\zeta - a)^{n+1} } d \zeta\right|  \le \frac{1}{2 \pi } \oint_{| \zeta - a | = \rho }\left| \frac{ f(\zeta) }{ (\zeta - a)^{n+1} }\right| d \zeta .$$
From here, I keep on deducing inegualities for $|a_n|$, until I find: $$|a_n| \le C_2 \frac{\log^{p} (\rho) }{ \rho^n } .$$
Now, I tried to break up the problem in who parts: $n$ 'small' ($ n < p$) and $n$ 'big' ($n \ge p$). For big $n$ I can take as big a $\rho$ as I want, so I rewrite the expression as follows: $$ |a_n| \le \lim_{\rho \to \infty}\left( \frac{\log(\rho) }{ \rho }\right)^p\left(\frac{1}{\rho}\right)^{n-\rho} =0 , $$ which is clear since both components of the product go to zero. So all $a_n$ must be zero, so $f$ is the zero function.
However, I am not sure how to treat the case $n < p$. Any ideas? 
Bonus question: is there an other, easier proof that $f$ is the zero function?
 A: 
Now,  I tried to break up the problem in who parts: $n$ 'small' ($n<p$) and $n$ 'big' ($n\ge p$). 

Good idea, poor execution. I understand the idea: when $n$ is big enough,  $\rho^n$ will grow faster than $\log^p \rho$, giving $a_n=0$ in the limit $\rho\to\infty$. But you were mistaken in thinking that $n$ needs to be $\ge p$ for this; in fact, any $n\ge 1$ is big enough. More precisely, 
$$\lim_{\rho\to\infty }\frac{\log^p\rho}{n^\epsilon}=0\quad \forall p>0\ \ \forall \epsilon>0$$
as you can check with L'Hospital. 
Now that we have $a_n=0$ for all $n\ge 1$, the function is shown to be constant. Also, at $z=0$ the given inequality $|f(z)| \le C \log^{p} (|z| + 1)$ turns into $|f(0)|\le 0$. Thus, $f$ is identically zero.
A: Your work shows that $f(z)$ is a polynomial with degree less than $p$. So you can see that $f(z)\equiv C$ by using argument by contradictory if $f(z)$ is a polynomial since otherwise
$$ \lim_{|z|\to\infty}\frac{|f(z)|}{\log^p(|z|+1)}=\infty$$
for any polynomial $f(z)$.
