Given the recurrence relation:
$p(n,k)=\frac{p(n-1,k-1)}{n} + \frac{(n-1)p(n-1,k)}{n}$ for $n \geq 2$
$p(1,k)=\delta_{k,0}$, $~ p_{n,k} = 0$ for $k<0$
I want to deduce that my recurrence satiesfies a similar recurrence as the unsigned stirling numbers of first kind:
${n+1 \brack k} = n {n \brack k} + {n \brack k-1}$ with ${0 \brack 0}=1$ and ${n \brack 0} = {0 \brack n} = 0$ for $k>0, n>0$
But as obvious I got problems with the intial values, because ${1 \brack k} = \delta_{1,k}$. I also tried to adjust the stirling numbers so it fits the initial values of my recurrence, but then I get problems with recurrence at ${2 \brack 1}$, which should be 1.
Is there a way to fix this or should I try another approach?
Let $a(n,k)=n!p(n,k)$. Then $a(1,0)=1$, $a(1,k)=0$ for $k>1$, $a(n,k)=0$ for $k<0$, and
$$a(n,k)=a(n-1,k-1)+(n-1)a(n-1,k)\;.$$
Here are a few values, with zeroes above the diagonal suppressed:
$$\begin{array}{c|cc} n\backslash k&0&1&2&3&4\\ \hline 1&1\\ 2&1&1\\ 3&2&3&1\\ 4&6&11&6&1\\ 5&24&50&35&10&1 \end{array}$$
For the unsigned Stirling numbers of the first kind we have the same recurrence,
$${n\brack k}={{n-1}\brack{k-1}}+(n-1){{n-1}\brack k}\;,$$
with initial values ${0\brack 0}=1$, ${n\brack 0}=0$ for $n>0$, ${0\brack k}=0$ for $k>0$, and ${n\brack k}=0$ for $k<0$.
Here are a few values, with zeroes above the diagonal suppressed:
$$\begin{array}{c|cc} n\backslash k&0&1&2&3&4\\ \hline 0&1\\ 1&0&1\\ 2&0&1&1\\ 3&0&2&3&1\\ 4&0&6&11&6&1\\ 5&0&24&50&35&10&1 \end{array}$$
The relationship is visually obvious:
$$a(n,k)={n\brack{k+1}}\tag{1}$$
for $n\ge 1$ and $k\ge 0$. And since $a(1,k)={1\brack{k+1}}$ for $k\ge -1$ and $a(n,-1)={n\brack 0}$ for $n\ge 1$, the fact that the $a(n,k)$ satisfy the same recurrence as the Stirling numbers of the first kind ensures that $(1)$ does indeed hold for $n\ge 1$ and $k\ge 0$. It follows that
$$p(n,k)=\frac1{n!}{n\brack{k+1}}\;.$$