Exponential generating functions (EGFs) can be used to represent different classes of connected undirected graphs. Composition of EGFs has the following interpretation.
If $F(x)$ represents graphs of type I and $G(x)$ represents graphs of type II, then $F(G(x))$ represents the graphs that can be constructed by first constructing a graph of type I and then replacing each of its vertices with a graph of type II.
It is straightforward to prove this in terms of a formula. Let $F(x) = \sum_{n \geq 0} \frac{f_n}{n!} x^n$ and $G(x) = \sum_{n \geq 1} \frac{g_n}{n!} x^n$. Then
$$F(G(x)) = \sum_{n \geq 0}\frac{1}{n!}\left(\sum_{k \leq n} \frac{f_k}{k!} \sum_{\begin{matrix}n_1, \ldots, n_k \,\geq\, 1\\n_1 + \cdots + n_k \,=\, n\end{matrix}} \binom{n}{n_1, \ldots, n_k} g_{n_1} \cdots g_{n_k}\right) x^n \tag{1}\label{a}$$
Moreover, after studying the formula for a moment, one observes that it expresses the interpretation of $F(G(x))$ mentioned above.
Is there is there a conceptual proof that does this in reverse? That is, can we prove that $F(G(x))$ has the boxed interpretation above without deriving the formula first algebraically? If so, then the formula \ref{a} for $F(G(x))$ would be immediate and hopefully such an argument would also provide some intuition into why the composition of two EGFs has such a nice interpretation.
Here is a straightforward but unenlightening derivation of the formula for \ref{a}. We start by considering two ordinary generating functions (OGFs) $A(x) = \sum_{n \geq 0} a_n x^n$ and $B(x) = \sum_{n \geq 1} b_n x^n$. Then
$$(B(x))^k = \sum_{n \geq k} \sum_{\begin{matrix}n_1, \ldots, n_k \,\geq\, 1\\n_1 + \cdots + n_k \,=\, n\end{matrix}} b_{n_1} \cdots b_{n_k} x^n$$
Therefore,
\begin{align} A(B(x)) & = \sum_{k \geq 0} a_k \sum_{n \geq k} \sum_{\begin{matrix}n_1, \ldots, n_k \,\geq\, 1\\n_1 + \cdots + n_k \,=\, n\end{matrix}} b_{n_1} \cdots b_{n_k} x^n \\ {} & = \sum_{k, n \geq 0} \;\; [n \geq k] \; a_k \sum_{\begin{matrix}n_1, \ldots, n_k \,\geq\, 1\\n_1 + \cdots + n_k \,=\, n\end{matrix}} b_{n_1} \cdots b_{n_k} x^n \\ {} & = \sum_{n \geq 0}\left(\sum_{k \leq n} a_k \sum_{\begin{matrix}n_1, \ldots, n_k \,\geq\, 1\\n_1 + \cdots + n_k \,=\, n\end{matrix}} b_{n_1} \cdots b_{n_k} \right) x^n \label{b} \tag{2} \end{align}
Now, consider the EGFs $F(x) = \sum_{n \geq 0} \frac{f_n}{n!} x^n$ and $G(x) = \sum_{n \geq 1} \frac{g_n}{n!} x^n$. Then by applying \ref{b}, we see that
\begin{align} F(G(x)) & = \sum_{n \geq 0}\left(\sum_{k \leq n} \frac{f_k}{k!} \sum_{\begin{matrix}n_1, \ldots, n_k \,\geq\, 1\\n_1 + \cdots + n_k \,=\, n\end{matrix}} \frac{g_{n_1}}{n_1!} \cdots \frac{g_{n_k}}{n_k!}\right) x^n\\ {} & = \sum_{n \geq 0}\frac{1}{n!}\left(\sum_{k \leq n} \frac{f_k}{k!} \sum_{\begin{matrix}n_1, \ldots, n_k \,\geq\, 1\\n_1 + \cdots + n_k \,=\, n\end{matrix}} \binom{n}{n_1, \ldots, n_k} g_{n_1} \cdots g_{n_k}\right) x^n \end{align}
which is precisely the formula \ref{a}.