TITLE: Stirling number of the first kind: Proof of Recursion formula
QUESTION [11 upvotes]: I want to prove this recursion formula for Stirling numbers of the first kind:
$$s_{n+1,k+1} = \sum_{i=k}^{n} \binom{i}{k} s_{n,i}$$
But I lack a useful idea.  Perhaps someone could inspire me?
Kind regards.

REPLY [4 votes]: This should go as comment, not as an answer(it doesn't provide additional proof), but is too long for the comment-box and may be interesting anyway. 
Your given relation describes a shift of the matrix of Stirlingnumbers 1st kind by a postmultiplikation of the Pascalmatrix:
$ S1_0 * P_0 = S1_1 $ where the index indicated a diagonal downshifting. See the matrices $S1_0, P_0,S1_1$ below:
$$ \small
\begin{matrix} 
  &  &  & S1_0 &  &  & | &  &  &  & P_0 &  &  & | &  &  & S1_1 &  &  &  \\
 1 & . & . & . & . & . & | & 1 & . & . & . & . & . & | & 1 & . & . & . & . & . \\
 -1 & 1 & . & . & . & . & | & 1 & 1 & . & . & . & . & | & . & 1 & . & . & . & . \\
 2 & -3 & 1 & . & . & . & | & 1 & 2 & 1 & . & . & . & | & . & -1 & 1 & . & . & . \\
 -6 & 11 & -6 & 1 & . & . & | & 1 & 3 & 3 & 1 & . & . & | & . & 2 & -3 & 1 & . & . \\
 24 & -50 & 35 & -10 & 1 & . & | & 1 & 4 & 6 & 4 & 1 & . & | & . & -6 & 11 & -6 & 1 & . \\
 -120 & 274 & -225 & 85 & -15 & 1 & | & 1 & 5 & 10 & 10 & 5 & 1 & | & . & 24 & -50 & 35 & -10 & 1
 \end{matrix}
$$
This can be iterated, so $ S1_{k+1} = S_k * P_k $      
Then the top-left-part of the result becomes -with increasing k - the identity and we can formulate the limit where $n \to \infty$ which extends to the identitymatrix. But this means, that that product of the down-shifted Pascalmatrices equals the inverse of $S1$ (up to dimension n) and on the other hand, the inverse is just the matrix of Stirlingnumbers 2'nd kind. So we get, denoting the top left segment of that matrix $S2$ up to dimension $n$ as  $$S2 = \prod_{k=0}^n P_k $$   and the three toplevel segments are  $P, P*P_1 , P*P_1*P_2 $ and the third of the three matrices below has its three left columns identical to the matrix of the Stirlingnumbers 2nd kind:
$$ \small
\begin{matrix} 
  &  &  & P_0 &  &  & | &  &  &  & P_0* &P_1  &  & | &  & P_0* & P_1* & P_2 &  &  \\
 1 & . & . & . & . & . & | & 1 & . & . & . & . & . & | & 1 & . & . & . & . & . \\
 1 & 1 & . & . & . & . & | & 1 & 1 & . & . & . & . & | & 1 & 1 & . & . & . & . \\
 1 & 2 & 1 & . & . & . & | & 1 & 3 & 1 & . & . & . & | & 1 & 3 & 1 & . & . & . \\
 1 & 3 & 3 & 1 & . & . & | & 1 & 7 & 5 & 1 & . & . & | & 1 & 7 & 6 & 1 & . & . \\
 1 & 4 & 6 & 4 & 1 & . & | & 1 & 15 & 17 & 7 & 1 & . & | & 1 & 15 & 25 & 9 & 1 & . \\
 1 & 5 & 10 & 10 & 5 & 1 & | & 1 & 31 & 49 & 31 & 9 & 1 & | & 1 & 31 & 90 & 52 & 12 & 1
 \end{matrix} $$
The both product-representations involving the shifted Pascalmatrices give then a nice generalization of your initial identity to (arbitrarily deep) nested products/sums of binomials for the Stirlingnumbers 1st and 2nd kind (a nice, inexhaustible, resource for homework-assignements, btw ;-) )
[update] I tried to relate to Mike Spivey's second identity. Surprise, we get the Bernoulli-numbers, scaled by factorials instead of binomials.
Denote the diagonalmatrix of factorials $F=diagonal(0!,1!,2!,...)$ and $f=F^{-1}$ and then 
the factorially scaled matrix of Stirlingnumbers 1st kind, whose c'th column provides the coefficients for the powerseries for $\ln(1+x)^c$, as $fS1F = f* S1 * F $ (this is a similarity scaling of $S1$ )  and the matrix of factorially scaled Bernoulli-numbers as $fBF$ then we get an inverse col/row shift by $ fS1F_1 * fBF = fS1F_0 $
$$ \small \begin{matrix} 
  &  &  & fS1F_1 &  &  & | &  &  &  & fBF &  &  & | &  &  & fS1F_0 &  &  &  \\
 1 & 0 & 0 & 0 & 0 & 0 & | & 1 & 0 & 0 & 0 & 0 & 0 & | & 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & 0 & | & -1/2 & 1 & 0 & 0 & 0 & 0 & | & -1/2 & 1 & 0 & 0 & 0 & 0 \\
 0 & -1/2 & 1 & 0 & 0 & 0 & | & 1/12 & -1/2 & 1 & 0 & 0 & 0 & | & 1/3 & -1 & 1 & 0 & 0 & 0 \\
 0 & 1/3 & -1 & 1 & 0 & 0 & | & 0 & 1/12 & -1/2 & 1 & 0 & 0 & | & -1/4 & 11/12 & -3/2 & 1 & 0 & 0 \\
 0 & -1/4 & 11/12 & -3/2 & 1 & 0 & | & -1/720 & 0 & 1/12 & -1/2 & 1 & 0 & | & 1/5 & -5/6 & 7/4 & -2 & 1 & 0 \\
 0 & 1/5 & -5/6 & 7/4 & -2 & 1 & | & 0 & -1/720 & 0 & 1/12 & -1/2 & 1 & | & -1/6 & 137/180 & -15/8 & 17/6 & -5/2 & 1
 \end{matrix}
 $$