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} $\square$