Limit of integer powers of a complex quantity 
Let $\varepsilon: \mathbb R\to \mathbb C$ be a function continuous at $0$ with $\varepsilon(0)=0$ and $t\in \mathbb R$.
Prove that $\displaystyle \lim_n \left(1-\frac{t}n + \frac{1}{n}\varepsilon\left(\frac{1}{n}\right)\right)^n=e^{-t}$ where the limit is taken on integers.

In other words, I want to prove that $\displaystyle \lim_n \left(1-\frac{t}n + o\left(\frac1n\right)\right)^n=e^{-t}$, where the error term $o(\frac1n)$ might be complex.
I know how to solve this when all terms are real: it suffices to write $$\left(1-\frac{t}n + \frac{1}{n}\varepsilon\left(\frac{1}{n}\right)\right)^n = \exp\left(n\log\left(1-\frac{t}n + o\left(\frac{1}{n}\right)\right)\right)$$
and perform asymptotic expansions.

My problem is that I have very limited knowledge of the complex logarithm. I'd like to see a formal proof that uses complex logarithm or other methods.

 A: The proof follows the same line of reasoning as the proof for real variables.  By definition, $z^c=e^{c\log(z)}$, where $\log(z)$ is the multi-valued complex logarithm.  
Note that $\log(z)=\log(|z|)+i\arg(z)$, where the argument of $z$ can be expressed as $\arg(z)=\text{Arg}(z)+2n\pi$ for $n\in \mathbb{Z}$ and $-\pi<\text{Arg}(z)\le \pi$ is the principal argument of $z$.
We can write
$$\left(1-\frac tn-\frac1n \epsilon\left(\frac1n\right)\right)^n=e^{n\log\left(1-\frac tn-\frac1n \epsilon\left(\frac1n\right)\right)}$$
The complex function $\log(1-z)$ is holomorphic for $z\in \mathbb{C}\setminus [1,\infty)$ (i.e., in the cut plane).  It can be written, therefore, in terms of its Taylor series (See here)
$$\log(1-z)=-\sum_{k=1}^\infty\frac{(-1)^kz^k}{k}$$
for $|z|<1$.  This is the expansion of $\log(1-z)$ on the principal branch.  The expansion on other branches are equal $\mod 2\pi$.
We fix $t$ and take $n$ so large that $\left|\frac tn+\frac1n \epsilon\left(\frac1n\right)\right|<1$.  Then, we can write
$$\log\left(1-\frac tn-\frac1n \epsilon\left(\frac1n\right)\right)=-\frac tn+o\left(\frac1n\right)$$
Can you finish it now?
A: I've found an elementary way that resorts to real methods only.
Write $$\begin{align}
\left(1-\frac{t}n + \frac{1}{n}\varepsilon\left(\frac{1}{n}\right)\right)^n&= \sum_{k=0}^n \binom nk \left(1-\frac{t}n  \right)^k \left( \frac{1}{n}\varepsilon\left(\frac{1}{n}\right) \right)^{n-k} \\
&= \underbrace{\left(1-\frac{t}n  \right)^n}_{\in \mathbb R} + \sum_{k=0}^{n-1} \binom nk \left(1-\frac{t}n  \right)^k \left( \frac{1}{n}\varepsilon\left(\frac{1}{n}\right) \right)^{n-k}
 \end{align}$$
It suffices to prove that $\displaystyle \sum_{k=0}^{n-1} \binom nk \left(1-\frac{t}n  \right)^k \left( \frac{1}{n}\varepsilon\left(\frac{1}{n}\right) \right)^{n-k} \xrightarrow[n\to\infty]{}0$ 
Choose $n$ big enough so that $1-\frac{t}n\geq 0$ and note that $$\left|\sum_{k=0}^{n-1} \binom nk \left(1-\frac{t}n  \right)^k \left( \frac{1}{n}\varepsilon\left(\frac{1}{n}\right) \right)^{n-k} \right|\leq \sum_{k=0}^{n-1} \binom nk \left(1-\frac{t}n  \right)^k \left( \frac{1}{n}\left|\varepsilon\left(\frac{1}{n}\right) \right| \right)^{n-k} $$
The sum on the right involves only real quantities, and in fact is equal to
$$\begin{align}\left(1-\frac{t}n + \frac{1}{n}\left|\varepsilon\left(\frac{1}{n}\right)\right|\right)^n - \left(1-\frac{t}n  \right)^n&=\left(1-\frac{t}n + \underbrace{o\left(\frac 1n \right)}_{\in \mathbb R}\right)^n - \left(1-\frac{t}n  \right)^n \\
&=o(1) \end{align}$$
