How to prove that $\lim \limits_{n \to \infty}(1+a_n)^n=e^p$ when $a_n \to 0$ and $n a_n \to p$? Assume that $(a_n)$ is a sequence of complex numbers and $p \in \mathbb{C}$.
How to prove, maybe without complex logarithm if possible, that:
If $a_n \rightarrow 0$ and $n a_n \rightarrow p$ then $(1+a_n)^n\to e^p$.
Such theorem is used for example in probability in proofs of some limits theorems.

Start from 
$$
(1+a_n)^n= \sum_{k=0}^{+\infty} c_{k,n}b_{k,n},\quad\text{with}\quad c_{k,n}=[k\leqslant n]\cdot \frac{n!}{n^k(n-k)!},\quad b_{k,n}=\frac{(na_n)^k}{k!}
$$
and try to apply Lebesgue dominated convergence theorem. To this end, note that, on the one hand,  $|c_{k,n}|\leqslant1$, and on the other hand, $na_n\to p$, hence there exists a finite $A$ such that $|na_n|\leqslant A$ for every $n$, in particular $|b_{k,n}|\leqslant\dfrac{A^k}{k!}$. This yields
$$
|c_{k,n}b_{k,n}|\leqslant\frac{A^k}{k!}\quad\text{with}\quad \sum_{k=0}^{+\infty}\frac{A^k}{k!}=\mathrm e^A\quad\text{finite},
$$
and the conditions of dominated convergence are met. Now, for every fixed $k$, $c_{k,n}\to c_k=1$ and $b_{k,n}\to b_k=\dfrac{p^k}{k!}$ when $n\to\infty$, hence
$$
\lim\limits_{n\to\infty}(1+a_n)^n=\sum_{k=0}^{+\infty}c_kb_k=\sum_{k=0}^{+\infty}\frac{p^k}{k!}=\mathrm e^p.
$$
