probability distributions - Binomial-Poisson limit

I want to show that if $Z_n$ has the binomial distribution with parameters $n$ and $\lambda/n$ with $\lambda$ fixed, then $Z_n $ converges in distribution to the Poisson distribution, parameter $\lambda$ as $n\rightarrow \infty$. How do I do this using characteristic functions?

Edit: i think the characteristic function of the binomial distribution is $(pe^{it}+(1-p))^n$ and that of the Poisson is $e^{\lambda(e^{it}-1)}$, but i dont know which limit to take.

Answer

So the characteristic function of $\text{B}(n,\lambda/n)$ is
$$ ((1-\lambda/n)+\lambda/n e^{it})^{n} = \left( 1 + \frac{1}{n} \lambda\left( e^{it}-1 \right) \right)^n. $$
Now use that
$$ \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x. $$
Then the convergence and uniqueness theorems for characteristic functions imply that the distribution is $\text{Po}(\lambda)$.