A translation of to domain $\Omega$ by $a$ and changing $\phi$ to $\phi-a$ clearly doesn't affect this computation at all, so we assume that $a = 0$. This is just a convenience; the computation just looks cleaner this way, and you can probably work it out just fine in the generality you posed by mimicking what occurs below.
$\phi$ has a power series expansion in some ball $B(0,R) \subset \Omega$ whose closure is contained in $\Omega$ as well. It looks like
$$\phi(z) = z + a_nz^n + \mbox{higher order terms} = z +a_nz^n +O(z^{n+1}).$$
Assume that the terms $a_2,\dots, a_{n-1}$ of the power series are all zero. We'll show $a_n$ must be zero as well. Also, I'm using big-O notation for convenience as well; if you don't like it you can just write 'higher order terms' instead and it won't change the reasoning.
To illustrate what is going on we compute the following easy case:
$$
\begin{align*}
\phi(\phi(z)) &= \phi(z) + a_n(\phi(z))^n + O(z^{n+1}) \\
&= z + a_n z^n + a_n\left( z + a_nz^n + O(z^{n+1})\right)^n + O(z^{n+1}) \\
&= z + 2 a_n z^n + O(z^{n+1})
\end{align*}
$$
In general (perhaps proven carefully using induction and the above reasoning) we have
$$\phi^{(m)}(z) = z + m a_n z^n + O(z^{n+1}).$$
Here, $\phi^{(m)}$ represents the $m$-fold composition of $\phi$.
The Cauchy estimates say that if $f$ is a holomorphic function in a neighborhood of some closed ball $B$ of radius $R$ and any $z$ in that ball we have
$$\left|\frac{d^n}{dz^n}f(z) \right|\leq \frac{n!\sup_{B}|f|}{R^n}.$$
Lets apply the Cauchy estimates to $\phi^{(m)}$ on $B(0,R)$. We can read off the $n$th derivative of this composition from the power series above to see that
$$\frac{d^n}{dz^n} \phi^{(m)}(z) = n!\cdot m \cdot a_n $$
Therefore the Cauchy estimates say that
$$|a_n| \leq\frac{ \sup_{B(0,R)}|\phi^{(m)}(z)|}{m\cdot R^n}$$
Since $n$ is fixed, $a_n$ and $R^n$ remain fixed quantities. Since $\phi: \Omega \rightarrow \Omega$, and $\Omega$ is a bounded domain, the supremum of $|\phi|$, and therefore of $|\phi^{(m)}|$ is uniformily (and most importantly, independently of $m$) bounded by some constant $C$, which corresponds to the norm of the point farthest from the origin in the domain. Rearranging everything, we have
$$|a_n| \leq \frac{K}{m}$$
where $K$ is some fixed real number independent of $m$. Since $m$ can be chosen arbitrarily large, we have $a_n = 0$.
We have shown there can be no higher order terms in the power series for $\phi$; $\phi$ must therefore be the identity.
