  In this short note we provide some algebraic identity with a proof exploiting
its probabilistic interpretation. We show several consequences of the identity,
in particular we obtain a new representation of a Stirling number of second
kind,
  $$ S(n,d)={1\over d!} \sum_{1\leq j_1<j_2<\ldots<j_{d-1}< n}
1\cdot2^{j_{d-1}-j_{d-2}}\cdots d^{j_1}$$
  for integers $n\geq d$. Relating this to other known formula for $S(n,d)$ we
also obtain
  $$ \sum_{1\leq j_1\leq j_2\leq \cdots\leq j_{n-d}\leq d} j_1j_2\ldots,j_{n-d}
=d! \sum_{1\leq j_1<j_2<\ldots<j_{d-1}< n} 1\cdot2^{j_{d-1}-j_{d-2}}\cdots
d^{j_1}.$$
  As a side effect, we have new proof of a known result stating that for any
integer $d\in\mathbb{N}$ and any $x\in\mathbb{R}$ equality
  $$\sum_{r=0}^d (-1)^r{d\choose r}(x-r)^d=d!$$ holds. This is a special case
of the presented identity.
