= Solution
Using the <probability mass function> of the <Poisson distribution> directly gives
$$
\boxed{\mathbb E[X^n]=\frac1e\sum_{j=0}^{\infty}\frac{j^n}{j!}.}
$$
For each fixed positive <integer> $n$, this <series> converges: the <ratio> of successive nonzero terms is $(1+1/j)^n/(j+1)$, which tends to zero. Equivalently, using the <polynomial identity> in (ii) and the <factorial moments> from (a)(iv) gives $\mathbb E[X^n]=\sum_{k=1}^nA(n,k)$; the extension of that identity to $X=0,1$ is justified in (iv).
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