Induction proof for stirling of first kind. I have to show that for every stirling number of the first kind $\forall n \geq 2 : s_{n,n-2} = \frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
I've started like this:
Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $\frac{1}{24}2(2-1)(2-2)(3*2-1) = \frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.
Now we assume $\forall n \in \mathbb{N} $ with $ n \geq 2$
that $s_{n,n-2} = \frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.
It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.
Now let $n \rightarrow n+1$.
$s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.
If we insert our assumption, that leaves us with
$s_{n+1,n-1} = \frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.
If we insert $n=n+1$ into $\frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
$\frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.
The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n \choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?

Induction step:
$\begin{aligned}s\left(n+1,n-1\right) & =ns\left(n,n-1\right)+s\left(n,n-2\right)\\
 & =\frac{1}{2}n^{2}\left(n-1\right)+\frac{1}{24}n\left(n-1\right)\left(n-2\right)\left(3n-1\right)\\
 & =\frac{1}{24}n\left(n-1\right)\left[12n+\left(n-2\right)\left(3n-1\right)\right]\\
 & =\frac{1}{24}n\left(n-1\right)\left(n+1\right)\left(3n+2\right)\\
 & =\frac{1}{24}\left(n+1\right)n\left(n-1\right)\left(3n+2\right)
\end{aligned}
$
In the second equality we use the equality $s(n,n-1)=\frac12n(n-1)$ and also the induction hypothesis.
