Let $g,i,k,n$ be non-negative integers, $p$ a prime number and ${n\brack {k}}$ (resp. ${n\brace k}$) be the Stirling numbers of first (resp. second) kind, such that: $$ \sum_{k\ge0} {{n}\brack {k}}x^k = \prod_{j=0}^{n-1}(x+j)$$ $$ \sum_{k\ge0} {{n}\brace {k}}\prod_{j=0}^{k-1}(x-j) = x^n$$
The following congruence, involving both kinds of Stirling numbers can be shown:
$$ \sum_{i \ge0}{{k+i-1}\choose{k-1}}{n\brack {k\cdot p-g+ i(p-1)}} \equiv {g\brace {k\cdot p-n}} \pmod p$$
Does someone know of any reference for this?