By definition of a
D2854: Poisson random natural number, we have
\begin{equation}
X
\overset{d}{=} \lim_{n \to \infty} \sum_{m = 1}^n \xi_{n, m}
\end{equation}
for a triangular array $\{ \{ \xi_{n, m} \}_{1 \leq m \leq n} \}_{n \geq 1}$ of Bernoulli random numbers $\xi_{n, m} \in \text{Bernoulli}(\theta_n)$ where $\theta_n = \min(\lambda / n, 1)$ for all positive integers $n \geq 1$. Fix now $n \geq 1$ such that $\lambda / n < 1$. Using result
R3200: Characteristic function of Bernoulli random boolean number, we can write
\begin{equation}
\mathbb{E} \left( e^{i t \xi_{n, m}} \right)
= \theta_n e^{i t} + 1 - \theta_n
= \frac{\lambda}{n} e^{i t} + 1 - \frac{\lambda}{n}
= 1 + \frac{\lambda (e^{i t} - 1)}{n}
\end{equation}
for all integers $1 \leq m \leq n$. Therefore, using the above result together result
R5114: Characteristic function for sum of independent random real numbers, we have
\begin{equation}
\begin{split}
\mathbb{E} \left( e^{i t \sum_{m = 1}^n \xi_{n, m}} \right)
= \prod_{m = 1}^n \mathbb{E} \left( e^{i t \xi_{n, m}} \right)
= \prod_{m = 1}^n \left( 1 + \frac{\lambda (e^{i t} - 1)}{n} \right)
= \left( 1 + \frac{\lambda (e^{i t} - 1)}{n} \right)^n
\end{split}
\end{equation}
Applying
R4123: Strong dominated convergence theorem for complex expectation, we then conclude
\begin{equation}
\begin{split}
\mathbb{E} \left( e^{i t X} \right)
& = \mathbb{E} \left( e^{i t \lim_{n \to \infty} \sum_{m = 1}^n \xi_{n, m}} \right) \\
& = \mathbb{E} \left( \lim_{n \to \infty} e^{i t \sum_{m = 1}^n \xi_{n, m}} \right) \\
& = \lim_{n \to \infty} \mathbb{E} \left( e^{i t \sum_{m = 1}^n \xi_{n, m}} \right) \\
& = \lim_{n \to \infty} \left( 1 + \frac{\lambda (e^{i t} - 1)}{n} \right)^n \\
& = e^{\theta (e^{i t} - 1)}
\end{split}
\end{equation}
$\square$
