How can I prove that $\lim_{n \to \infty}\left(1+c\cdot \left(\exp\left(\frac{it}{n}\right) - 1\right)\right)^n=\exp(i\cdot tc)$? I'm trying to prove some random variable converges to a constant $c$ and this limit showed up
$$\lim_{n \to \infty}\left(1+c\cdot \left(\exp\left(\frac{it}{n}\right) - 1\right)\right)^n$$
For the random variable in question to converge to $c$ it's necessary that this limit above be equal to $\exp(itc)$, which is the reason for the title. I tried using the classic:
$$\lim_{n \to \infty}\left(1+\frac{z}{n}\right)^n=e^z$$
but this did not help me. I'm thinking I'm gonna have to use some other known limit involving the exponential but I don't know how to do so. I appreciate any help!

Hints: The pricipal logarithm $Log \,z$ is  continuous on the disk $B(1,1)$ and it satisfies the property $\frac {Log \, z } {z-1} \to 1$ as $z \to 1$. The given limit is $e^{ \lim n Log \, [1+c((e^{it/n}-1))]}$. Use the fact that $n(e^{ix/n}-1) \to ix$ as $ n \to \infty$ to finish the proof.
