Define $$G(x)=\sum_{n\geq0}n!x^n$$ (we don't care if it converges).
A permutation is indecomposable if it cannot be cut into two non-empty parts so that everything before the cut is smaller than everything after the cut. For example, $21, 3142, 2413 $ and many other permutations are indecomposable, while $23165$ is decomposable, since we can cut it as $231|65$. Let $a_n$ denote the number of indecomposable permutations with $a_0 = 0$ and $A(x)= \sum_{n\geq 0}a_n x^n$. Express $A(x)$ in terms of $G(x)$.
I see how $n!$ and permutations are related but am struggling to produce anything resembling an answer.
A decomposable permutation corresponds to a pair of shorter permutations if you renumber the second part to start at $1$. So for a start, we can count the decomposable permutations using $(G-1)^2$ (where we need to subtract $1$ because the empty permutation shouldn't be included). But this overcounts permutations that can be decomposed in more than one way; e.g. $123$ is counted both as $12|3$ and as $1|23$. We can correct for this by subtracting $(G-1)^3$; but then we're undercounting e.g. $1234$, which can be decomposed in three ways and has been counted three times in $(G-1)^2$ and three times in $(G-1)^3$. So we need to add $(G-1)^4$ back in. Continuing in this manner, we can count the decomposable permutations using
$$ (G-1)^2-(G-1)^3+(G-1)^4-\cdots\;. $$
Indeed, a permutation that can be decomposed into $k$ "components" is counted $\sum_{m=2}^k(-1)^m\binom{k-1}{m-1}=1$ times.
To count the indecomposable permutations, we subtract this from $G$:
\begin{eqnarray*} &&G-(G-1)^2+(G-1)^3-(G-1)^4+\cdots\\ &=&1+(G-1)-(G-1)^2+(G-1)^3-(G-1)^4+\cdots\\ &=&1-\frac1G\;. \end{eqnarray*}