{c}
{title2=$B_n=e^{-1}\sum_{j=0}^\infty j^n/j!$}

= Bell number as a Poisson moment
{c}
{synonym}

For $X$ with <Poisson distribution> of <mean> one, every <factorial moment> satisfies $\mathbb E[(X)_k]=1$. The <polynomial> identity
$$
x^n=\sum_{k=0}^n S(n,k)(x)_k
$$
expresses powers in terms of <falling factorials>, where $S(n,k)$ are <Stirling numbers of the second kind>. Substituting $X$ and using <linearity of expectation> proves
$$
\mathbb E[X^n]=\sum_{k=0}^n S(n,k)=B_n.
$$
On 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.


 Dobiński's formula (source code)