$f$ is entire, show that the order of $f'$ is less than or equal to that of $f$ Let $f$ be an entire function of finite order. For entire functions $g$ of finite order, let $o_g$ denote the order of $g$ (recall that this order is defined as the infinum of all nonnegative constants $a$ so that $\exists r>0$ where $\rvert z\rvert > r$ implies that $\rvert g(z)\rvert < exp(\rvert z\rvert^a)$).

Problem that I am working on:
Show that $o_{f'}\leq o_f$.

Here is what I have so far:
I think that I am supposed to use Cauchy's integral formula and the fact that: $o_f = \frac{loglogM(r)}{log(r)}$, where $M(r):= M_f(r):= max_{\rvert z\rvert \leq r}\rvert f(z)\rvert$ (1). Notice how (1) gives that it suffices to show that eventually (for sufficiently large $r$) $M_{f'}(r)\leq M_f(r)$ (since log is an increasing function).
To get this, I had thought that Cauchy's integral formula would work, but I am not getting anywhere with it. Namely, if $z\in \mathbb{C}$ and $C_r = C_r(z)$ denotes (for all fixed $r>0$) the circle of center $z$ and radius $r$, we have:
$\rvert f'(z)\rvert = \rvert \frac{1!}{2\pi i}\int_{C_r}\frac{f(w)}{(w-z)^2}dw\rvert$
If we fix the radius of $z$ we are considering at $R$, we get for instance, for all $r>0$:
$M_{f'}(R)\leq \frac{2\pi}{2\pi}\frac{M(R+r)}{r} = \frac{M(R+r)}{r}$,
which is completely unhelpful (I would like $r$ to be sent to zero, but we can't really do that here).
Any help is appreciated (I have been goofing around with this for quite some time now, I also tried parameterizing before moving the absolute value inside the integral and estimating, but then I run into more or less the same issue).
 A: You already demonstrated that
$$
M_{f'}(R)\le  \frac{M(R+r)}{r}
$$
for $0 < r < R$, and that is helpful. In fact it suffices to use that
$$
 M_{f'}(R)\le M(R+1) \, .
$$
If $\rho(f) < a$ then $M(r) < \exp( r^a)$ for sufficiently large $r \ge r_0$, and therefore
$$
M_{f'}(r)\le \exp( (r+1)^a) 
$$
or
$$
\frac{ \log\log M_{f'}(r)}{\log(r)} \le a \frac{\log(r+1)}{\log(r)} 
= a \left( 1 + \frac{\log(1+1/r)}{\log(r)}\right)
\,.
$$
It follows that
$$
\rho(f') =  \limsup_{r \to \infty}\frac{ \log\log M_{f'}(r)}{\log(r)} \le a \, .
$$
This holds for all $a > \rho(f)$ and therefore is $\rho(f') \le \rho(f)$.
Remark: As Conrad said, one can also compute the order of an entire function $f(z) = \sum_{n=0}^\infty a_n z^z$ from its Taylor coefficients:
$$
\rho =\limsup _{n\to \infty }{\frac {n\ln n}{-\ln |a_{n}|}} \, .
$$
This shows that $f$ and $f'$ have in fact the same order.
A: Suppose $|f(z)|\le \exp(|z|^a)$ for $|z|>r_0$. For $|z|>2r_0+1$, Cauchy integral formula gives
$$ |f'(z)| \le C\int_{B(z,\epsilon)} \frac{|f(w)|}{|z-w|^2}|dz| \le \frac {C}{\epsilon} \sup_{v\in B(0,\epsilon)}\exp(|z+v|^a)$$
By taking $\epsilon\ll 1$, we see that
$$ |f'(z)|\le C \exp(C|z|^a)$$
for a large constant $C$. It follows that for any $\delta>0$,  by choosing a $r_0\gg1$,
$$ |f'(z)|\le  \exp(|z|^{a+\delta}).$$
So we have shown that if $o_f \le a$ then $o_{f'}\le a$. By taking an infimum over all $a$ such that $o_f\le a$, we obtain the result.
