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.

Eited
Too long for a comment, so i edited the first "real" solution.
$$(1+a_n)^n= \sum_{k=0}^n \frac{n!}{k! (n-k)!}a_n^k \,.$$
$$e^{na_n}=   \sum_{k=0}^n \frac{n^k}{k!}a_n^k + \sum_{k=n}^\infty \frac{n^k}{k!}a_n^k\,.$$
Thus,
$$ \left|(1+a_n)^n- e^{na_n}  \right| \leq \left| \sum_{k=0}^n \frac{n(n-1)..(n-k)-n^k}{k!}a_n^k  \right| +\left| \sum_{k=n}^\infty \frac{n^k}{k!}a_n^k\right|$$
It is obvious that the second term converges to $0$, you can use the continuity of the expenential+ uniform convergence of the Taylor series.
The first one is 
$$\left| \sum_{k=0}^n \frac{(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{k}{n})-1}{k!}(na_n)^k  \right|$$, should be eay to show it is convergent to 0. 
