Dobiński's formula ($B_n=e^{-1}{\sum }_{j=0}^{\infty }{j}^n/j!$)

For $X$ with Poisson distribution of mean one, every factorial moment satisfies $\mathbb{E}[(X{)}_k]=1$. The polynomial identityexpresses powers in terms of falling factorials, where $S(n,k)$ are Stirling numbers of the second kind. Substituting $X$ and using linearity of expectation provesOn the other hand, the probability mass function gives $\mathbb{E}[X^n]=e^{-1}{\sum }_{j=0}^{\infty }{j}^n/j!$. This proves Dobiński's formula and explains why a count of set partitions is a moment of a Poisson distribution. For example, $\mathbb{E}[X^3]=\mathbb{E}[(X{)}_3+3(X{)}_2+X]=1+3+1=5=B_3$. At $n=0$, take $j^0=1$ also for $j=0$, obtaining $B_0=1$. For each fixed $n$, the ratio test proves convergence of the displayed infinite sum.