= Solution
Both sides of (ii) are <polynomials> in the colour count, and they agree at every <integer> $x\ge2$. A nonzero <polynomial> has only finitely many roots, so this is a <polynomial identity> and holds also at $x=0,1$. It can therefore be evaluated at every possible value of the <Poisson random variable> $X$.
Take <expectations> and use the <factorial moments> from (a)(iv). The sum over $k$ is finite, so <linearity of expectation> gives
$$
\mathbb E[X^n]
=\sum_{k=1}^nA(n,k)\mathbb E[X(X-1)\cdots(X-k+1)]
=\sum_{k=1}^nA(n,k)=B(n).
$$
Combining this with (iii) proves <Dobiński's formula> for the <Bell numbers>:
$$
\boxed{B(n)=\frac1e\sum_{j=0}^{\infty}\frac{j^n}{j!}\qquad(n\ge1).}
$$
For completeness, the empty <set> has one <partition of a set>, giving $B(0)=1$; that case uses the empty-product conventions $0^0=1$ in this identity and a $k=0$ term in the partition sum.
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