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.