Proof that composition on $Aut(G)$ is associative

Question

Would it be correct to prove it the way below? I'm feeling somewhat ambiguous as to the correctness of my proof. I would appreciate your input.

Proof:

We need to prove that for $\phi$, $\psi$, $\rho \in Aut(G)$, $\phi(\psi\rho)=(\phi\psi)\rho$. Suppose $g\in G$, then:

(i) $[(\phi\psi)\rho](g)=\phi[\psi\rho(g)]=\phi[\psi(\rho(g))]=\phi(\psi(g_\rho))=\phi(g_{\rho\psi})=g_{\rho\psi\phi}$.

(ii)$[\phi(\psi\rho)](g)=\phi[(\psi\rho)(g)]=\phi[\psi(\rho(g))]=\phi(\psi(g_\rho))=\phi(g_{\rho\psi})=g_{\rho\psi\phi}$.

Thus $Aut(G)$ is associative.

Answer 1

It should be well known that composition of functions is always associative. If not, here is the proof:

Suppose $f:A\to B$, $g:B\to C$ and $h:C\to D$ are functions. For any $a\in A$ $$ ((h\circ g)\circ f)(a)=(h\circ g)(f(a))=h(g(f(a))) $$ and $$ (h\circ(g\circ f))(a)=h((g\circ f)(a))=h(g(f(a))). $$ The two expressions are equal.