Bounded spherical derivative implies finite order Let $f$ be an entire function. The Spherical Derivative $\rho(f)$ is defined by
$$\rho(f)(z):= \frac{|f'(z)|}{1+|f(z)|^2}.$$
A result from Clunie and Hayman states that if $\rho(f)$ is bounded, then $f$ is of exponential type. The proof uses the machinery of Nevanlinna's theory of value distribution.
My question is the following :
Is there an "elementary" proof that if $\rho(f)$ is bounded, then $f$ is of finite order?
(Note that this is a weaker result, since I'm only asking for finite order here).
Finite order means that there exists constants $K$ and $\alpha$ such that
$$|f(z)| \leq Ke^{|z|^\alpha}$$
for all $z$.
Motivation :
Motivation : I'm interested in this because it would lead to a quick proof of Picard's little theorem. Indeed, if there exists a non-constant entire function which omits $0$ and $1$, then it is possible to obtain (using normal families techniques) a non-constant entire function $f$ which omits $0$ and $1$ and that has bounded spherical derivative. Write $f=e^g$ for some entire function $g$. Since $f$ is of finite order, $g$ is a polynomial. But f does not take the value $1$, so g must be constant, a contradiction.
Any reference is welcome,
Malik
NOTE: This is a duplicate of a question on MathOverflow. I'm posting it here too because I did not get any answer or comment.
 A: I will sketch a proof. This is obtained by isolating the relevant parts of Nevanlinna theory, and I am following Nevanlinna's treatment of the Ahlfors-Shimizu characteristic (Chapter VI, Section 3 in his book). 
Let us assume without loss of generality that $f(0)=0$. Let $f^{\#}(z)$ denote the spherical derivative, as above. Set
$$A(r) := \int_{|z|<r} (f^{\#}(z))^2dxdy,$$
so $A(r)$ is the spherical area of the image of the disk of radius $r$. If $f^{\#}(z)$ is bounded, then 
$$A(r) = O(r^2).$$
An application of Green's theorem gives
$$4 A(r) = r\cdot \frac{d}{dr} \int_{0}^{2\pi} \log(1+|f(re^{i\theta})|^2)d\theta.$$
(I will leave it to you to check the details.)
We consider the expression
$$m(r) := \frac{1}{2\pi} \int_{0}^{2\pi}\log(1+f(re^{i\theta})|^2)d\theta.$$
By the above, we have 
$$\frac{d}{dr} m(r) = O(r), $$
and hence 
$$m(r) = O(r^2)$$
as $r\to\infty$.
Since $\log(1+|f|^2)>2\log|f|$, we thus have
$$\int_0^{2\pi} \log|f(re^{i\theta})|d\theta \leq \operatorname{const}\cdot r^2$$
for all sufficiently large $r$. By the Poisson integral formula for the harmonic function $\log|f|$, this implies that, for $|z|=r/2$,
$$\log|f(z)| \leq \operatorname{const}\cdot r^2=\operatorname{const}\cdot |z|^2.$$
So $f$ has finite order, as claimed.
