I am trying to understand the derivation of the Stirling Numbers from a difference table.
From my book:
Let $h_n = n^p$. The $0$-th diagonal of the difference table for $h_n$ has the form $c(p,0), c(p,1), c(p,2),\dots,c(p,p),0,0,\dots\;\;$.
This is where I am confused: what is $c(p,p)$? I at first thought they were the binomial coefficients, but upon computing the difference table for $h_n = n^4$:
$ \begin{array}{rrrrrrrrr} 0, & & 1, & & 16, & & 81, & & 256 \\ & 1, & & 15, & & 65, & & 175 \\ & & 14, & & 50, & & 110 \\ & & & 36, & & 60 \\ & & & & 24 \end{array} $
The $0$-th diagonal is clearly: $0, 1, 14, 36$, but $0 \ne \binom40$, $1 \ne\binom41$, etc.
In the end the Stirling Numbers of the Second Kind are derived to be: $S(p,k) = c(p,k)/k!$
The book Introductory Combinatorics by Brualdi (5th edition) can be found online here: filetosi.files.wordpress.com/2010/12/combiatoric.pdf. My questions about c(p,k) begin at pg. 281, halfway through the page.