Correctness of proof for finite sum of Stirling numbers of the first kind

Question

it is known that the finite sum of Stirling numbers of the first kind $s_{n,k}$ is $n!$ as defined below.

$$\sum_{k=0}^{n} s_{n,k} = n!\tag{1}$$

I attempt to prove this by induction, where the base case $n=1$ can be easily proven since we know that $\begin{bmatrix} n\\ n\end{bmatrix} = 1$ and $\begin{bmatrix} n\\ 0\end{bmatrix} = 0$.

Assuming equation $(1)$ holds, we move on now to prove the following:

$$\sum_{k=0}^{n+1} s_{n+1,k} = (n+1)!$$

Starting from the left hand side we have

$$\begin{align*} \sum_{k=0}^{n+1} s_{n+1,k} &= \sum_{k=0}^{n+1} s_{n,k-1} + ns_{n,k}\\\\ &= \sum_{k=0}^{n+1} s_{n,k-1} + \sum_{k=0}^{n+1} ns_{n,k}\\\\ &= n! + n\sum_{k=0}^{n+1} s_{n,k}\\\\ &= n! + n(n!)\\\\ &= (n+1)!\quad\square\quad\text{(since }s_{n,n+1}=0\text{)} \end{align*}$$

Is my proof as such correct and rigorous enough?

Answer 1

As often, it all depends on what you consider given. If you consider the recurrence relation as given, your proof looks OK. But that would only really make sense if you know how to prove that recurrence relation without appealing to the fact that $s_{n,k}$ counts the number of permutations of $n$ objects with $k$ cycles, because if you're going to use that fact, you don't need a proof by induction, since it almost immediately implies what you're trying to prove.