= Solution
A simple birth process with parameter $\lambda$ starting from one individual is the <Yule process>: when its population is $n$, it waits an exponential time of rate $n\lambda$ and then moves to $n+1$.
For $f(n)=n$, the <Markov jump-process generator> satisfies
$$
(Lf)(n)=n\lambda\bigl(f(n+1)-f(n)\bigr)=n\lambda.
$$
Therefore $\mu(t)=\mathbb E X_t$ obeys
$$
\mu'(t)=\lambda\mu(t),
\qquad \mu(0)=1.
$$
It follows, as recorded by the <mean population of a Yule process>, that
$$
\boxed{\mathbb E X_t=e^{\lambda t}.}
$$
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