This is a solution present in the book Mathematical Induction A powerful and elegant method of proof. ] I didn't quite understand the use of $L_n - nL_{n-1}$ in $(***)$
Problem 9.2.4. Prove that for any $n\in\mathbb{Z}, n\geq1$, one has
$$ \lim _{x \rightarrow 0} \frac{n ! x^{n}-\sin (x) \sin (2 x) \ldots \sin (n x)}{x^{n+2}}=\frac{n(2 n+1)}{36} \cdot n ! \text {. } $$
Solution. We shall prove that the limit is well defined for each $n \geq 1$ while we are finding the suitable recurrence relation.
Let $L_{n}$ be the corresponding limit to $n \geq 1$. For $n=1$, we have by repeated application of L'Hôpital's rule that
$$ L_{1}=\lim _{x \rightarrow 0} \frac{x-\sin (x)}{x^{3}}=\frac{1}{6} $$
Also, one has
$$ \begin{aligned} L_{n}-n L_{n-1} & =\lim _{x \rightarrow 0} \frac{n x \sin (x) \ldots \sin (x(n-1))-\sin (x) \sin (2 x) \ldots \sin (n x)}{x^{n+2}} (***) \\ & =\lim _{x \rightarrow 0} \frac{\sin (x)}{x} \cdot \frac{\sin (2 x)}{x} \ldots \frac{\sin ((n-1) x)}{x} \lim _{x \rightarrow 0} \frac{n x-\sin (n x)}{x^{3}} \\ & =(n-1) ! \lim _{x \rightarrow 0} \frac{n-n \cdot \cos (n x)}{3 x^{2}} \\ & =n ! \lim _{x \rightarrow 0} \frac{n \sin (n x)}{6 x} \\ & =n ! \cdot \frac{n^{2}}{6} . \end{aligned} $$
Therefore we have $L_{n}=n L_{n-1}+n ! \cdot \frac{n^{2}}{6}$.
We now prove by induction the statement
$$ P(n): \quad L_{n}=\frac{n(2 n+1)}{36} \cdot(n) ! $$
For $n=1$ we proved the result above, since $\frac{1 \cdot 3 \cdot 2 !}{36}=\frac{1}{6}$.
Assume now that $P(n-1)$ is true. From the above recurrence we have
$$ \begin{aligned} L_{n} & =n \cdot \frac{(n-1)(2 n-1) \cdot n !}{36}+\frac{n^{2} \cdot n !}{6}=\frac{n ! \cdot n}{36}[(n-1)(2 n-1)+6 n] \\ & =\frac{n ! \cdot n(n+1)(2 n+1)}{36}=\frac{n(2 n+1)}{36}(n+1) ! . \end{aligned} $$
This establishes the induction step and completes our proof.
That proof is flawed indeed, because assuming that $P(n-1)$ is true, what we get is $$L_n=n \cdot \frac{(n-1)(2 n-1) \cdot\color{red}{(n-1)!}}{36}+\frac{n^{2} \cdot n !}6,$$ not $$L_n=n \cdot \frac{(n-1)(2 n-1) \cdot n !}{36}+\frac{n^{2} \cdot n !}6. $$
Actually, the claim itself is wrong, as you can notice for instance for the base case $n=1$. The correct formula is $$L_n=\frac{n(2n+1)}{36}\color{red}{(n+1)!},$$ as can be either found directly, or proved by induction along the following lines: $L_1=\frac16=\frac{1 \cdot 3}{36}\color{red}{2!}$ and $$n\frac{(n-1)(2n-1)}{36}n!+n!\frac{n^2}6=\frac{n!n}{36}\left((n-1)(2n-1)+6n\right)\\=\frac{n(2n+1)}{36}(n+1)!.$$