I'm trying to see more clearly the reason behind the connection of the two following situations:

Situation 1: The signless Stirling numbers of the first kind, where $s(n,k)$ counts the number of permutations of $n$ elements that are comprised of a $k$ number of cycles.

Situation 2: Starting at $n$, choose a random integer in $0$ to $n-1$ (with equal likelihood). If you pick $0$, stop and say the process took 1 step. Otherwise repeat, counting the number of steps until $0$. Let $P(n,k)$ be the probability of starting at $n$ and taking $k$ steps to reach $0$. (You can think of this process as doing the Euclidean Algorithm where the remainders are always taken between $0$ and one less than the previous remainder.)

Working through a recursive computation for $P(n,k)$ leads to the fact that $$P(n,k) = \frac{s(n,k)}{n!}$$ (since $n!P(n,k)$ satisfies the same recursive formula as the signless Stirling numbers of the first kind. I can show the details if you think it'd help.)

My question: What is the relationship/reason behind the above equality? How does $s(n,k)$ count descents of length $k$ and/or vice versa?

For example: Take $n=4$, and $k=2$. Then $s(4,2) = 11$ coming from the 11 permutations: $(12)(34)$, $(13)(24)$, $(14)(23)$, $(123)(4)$, $(132)(4)$, $(124)(3)$, $(142)(3)$, $(134)(2)$, $(143)(2)$, $(1)(234)$, $(1)(243)$.

On the other side, $P(4,2)$ can be computed as follows: The 2-step decent $4-3-0$ happens with probability $\frac{1}{4}\cdot\frac{1}{3} = \frac{2}{24}$. The decent $4-2-0$ happens with probability $\frac{1}{4}\cdot\frac{1}{2} = \frac{3}{24}$. and $4-1-0$ has probability $\frac{1}{4} = \frac{6}{24}$. So $P(4,2) = \frac{11}{24}$.

So what's the explicit matching/counting going on between the $k$-cycle permutations and the $k$-step descents?