Show $f_n = f \circ f \circ \dots \circ f \longrightarrow 0$ uniformly on compact sets I am seeking help on a complex analysis qualifying exam problem.  

Let $D$ be a bounded open connected subset of $\mathbf{C}$ containing $0$ and let $f \colon D \to D$ be an analytic function satisfying $f(0) = 0$ and $\left| f^\prime \right|(0) < 1$.  Define $f_n = f \circ f \circ \dots \circ f$ ($n$ times).  Prove that $f_n \longrightarrow 0$ uniformly on compact sets.  

The hint is to start locally around zero.  I was able to prove that there exists a neighborhood $U$ contained within the radius of convergence of $f$ about $0$ such that $f_n \longrightarrow 0$ uniformly on compact sets contained in $U$.  I note that the proof did not use the boundedness of $D$.  
I am having trouble extending the result to the entirety of $D$.  Perhaps I am supposed to exploit the connectedness of $D$, considering something like $E = \left\{ z \in D \colon \text{ the result is true locally around } z \right\}$ and showing this set is open and closed.  If this is true, then for an arbitrary compact subset of $D$ we can take a finite cover applying the local result and be done with it.  It is obvious that $E$ is open, but I am having trouble showing it is closed.  
I don't have an idea where the boundedness of $D$ comes into play...
Many thanks in advance for your help.
 A: Assume we have proved there exists $U \ni 0$ open on which $f_n \longrightarrow 0$ uniformly on compact sets.  (This is not too hard.  Start with $U$ small enough so that the power series for $f$ at $0$ converges on $U$.  Then by assumption, for $z \in U$ we have $$ \left| f(z) \right| = \left|z\right| \left| f^\prime(0) + c_2 z + \dots \right|.  $$  Since $f^\prime(0) < 1$, upon shrinking $U$ further, by continuity there exists $\alpha < 1$ such that $$ \left| f(z) \right| < \alpha |z|. $$  If $K \subset U$ is compact, then $$ \left| f_n(z) \right| \leq \alpha^n \sup_{w \in K} \left| w \right| \longrightarrow 0.)$$  
Now, since $D$ is bounded and $f$ maps into $D$, $f_n(z)$ is uniformly bounded, hence by the Arzelà-Ascoli theorem for analytic functions (i.e. Montel's theorem) there is a subsequence $f_{n_k}$ that converges uniformly on compact sets to some analytic function $f$.  In particular, since $f_n \longrightarrow 0$ on $U$, $f(U) = 0$, hence $f = 0$.    
Fix $\epsilon > 0$ and let $K$ be a compact subset of $D$.  By what we just showed, there exists $N$ such that $\left| f_N(K) \right| \leq \epsilon$.  Take $\epsilon$ sufficiently small so that $K^\prime \subset U$, where $K^\prime = f_N(K)$.  Since $f_n \longrightarrow 0$ uniformly on $K^\prime$, take $M$ so that for all $n \geq M$, $\left| f_M(K^\prime) \right| \leq \epsilon$.  But $f_M \circ f_N = f_{N + M}$, hence for all $n \geq N + M$ we conclude $$ \left| f_n(K) \right| \leq \epsilon. $$  
