In Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick it is mentioned that the generating function of Lah numbers (called fragmented permutations) is \begin{equation} \displaystyle\sum_{n=0}^{\infty}L(n)\frac{x^n}{n!}=e^{x/(1-x)} \end{equation}
but in this questionthey arrive at another result using even another notation ($L (n, k)$).
I understand that $L(n)\neq L(n,k)$ since they have different recurrence relationships. my question is, knowing that the recurrence relationship of $L(n)$ is
\begin{equation} L(n+1)=(2n+1)L(n)-(n^2-n)L(n-1); n\geq 1 \end{equation}
How to can prove
\begin{equation} \displaystyle\sum_{n=0}^{\infty}L(n)\frac{x^n}{n!}=e^{x/(1-x)} \end{equation}
We can certainly prove the recurrence from the EGF. With $L_n = n! [z^n] \exp(z/(1-z))$ which BTW is the combinatorial class (compare to Stirling numbers)
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\textsc{SEQ}_{\ge 1}(\mathcal{Z}))$$
we obtain by differentiating
$$L(z) = \exp(z/(1-z))$$
that
$$n! [z^n] L'(z) = L_{n+1} = n! [z^n] L(z) \frac{1}{(1-z)^2} \\ = n! \sum_{q=0}^n [z^q] L(z) [z^{n-q}] \frac{1}{(1-z)^2} = n! \sum_{q=0}^n \frac{1}{q!} L_q (n-q+1) \\ = (n+1)! \sum_{q=0}^n \frac{1}{q!} L_q - n! \sum_{q=1}^n \frac{1}{(q-1)!} L_q.$$
This also yields
$$(n+1) L_{n+1} = - (n+1)! \sum_{q=0}^n \frac{1}{q!} L_q + (n+2)! \sum_{q=0}^n \frac{1}{q!} L_q - (n+1)! \sum_{q=1}^n \frac{1}{(q-1)!} L_q.$$
Subtract from
$$L_{n+2} = (n+2)! \sum_{q=0}^{n+1} \frac{1}{q!} L_q - (n+1)! \sum_{q=1}^{n+1} \frac{1}{(q-1)!} L_q$$
to get
$$L_{n+2} = (n+1) L_{n+1} + (n+1)! \sum_{q=0}^n \frac{1}{q!} L_q + (n+2) L_{n+1} - (n+1) L_{n+1} \\ = (n+2) L_{n+1} + (n+1)! \sum_{q=0}^n \frac{1}{q!} L_q.$$
This yields
$$(n+2) L_{n+2} = (n+2)^2 L_{n+1} + (n+2)! \sum_{q=0}^n \frac{1}{q!} L_q$$
as well as
$$L_{n+3} = (n+3) L_{n+2} + (n+2)! \sum_{q=0}^{n+1} \frac{1}{q!} L_q.$$
Subtract one more time to get
$$L_{n+3} = (2n+5) L_{n+2} - (n+2)^2 L_{n+1} + (n+2) L_{n+1} \\ = (2n+5) L_{n+2} - (n+2)(n+1) L_{n+1}$$
Setting $n$ to $n-2$ we get
$$L_{n+1} = (2n+1) L_n - n(n-1) L_{n-1}$$
as claimed.