Proof: $\lim \limits_{n\to\infty}(1+\frac{z}{n})^n = \exp(z)$ I define $\exp: \mathbb C \to \mathbb C$ as $z \mapsto \sum \limits_ {k=0}^{\infty}\frac{z^k}{k!}$. I would like to show that $\lim \limits_{n\to\infty}(1+\frac{z}{n})^n = \exp(z)$. I have a proof for the case $z \in \mathbb R$, but the proof assumes that $\lim \limits_ {n\to\infty}(1+\frac{z}{n})^n$ exists, which is easy to see if $z \in \mathbb R$, but not that easy (for me) if $z \in \mathbb C$. 
Would be good to have a proof which does not use derivatives or integrals.    
 A: I give another approach based on same technique as mentioned in this answer, but a lot simpler. Note that using the series definition of $\exp(z)$ you can prove that $$\exp(z + w) = \exp(z)\exp(w)$$ and therefore $\exp(z)\exp(-z) = 1$ so that $\exp(z) \neq 0$ for all $z \in \mathbb{C}$.
Now consider the sequence $$a_{n} = \dfrac{1 + \dfrac{z}{n}}{\exp\left(\dfrac{z}{n}\right)} = \left(1 + \frac{z}{n}\right)\exp(-z/n) = 1 - \frac{z^{2}}{n^{2}} + \dots$$ where $\dots$ represent terms with higher powers of $z/n$ so that we can write $$a_{n} = 1 - \frac{z^{2}}{n^{2}} + o(1/n^{2})$$ and therefore $n(a_{n} - 1) \to 0$ as $n \to \infty$. It follows from the theorem mentioned in the linked answer that $a_{n}^{n} \to 1$ and hence $$\lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n} = \exp(z)$$

The simplicity of this approach is because of the series representation of $\exp(z)$. In the linked answer the series for $\exp(z)$ is not used and instead I prove that if $z = x + iy$ then $$\lim_{n \to \infty}\left(1 + \frac{z}{n}\right)^{n} = e^{x}(\cos y + i\sin y)$$ where $e^{x}$ is defined by $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ Combining the current answer and the linked answer we see that for $z = x + iy$ we have $$\exp(z) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!} = e^{x}(\cos y + i\sin y)$$ and hence putting $x = 0$ and comparing real and imaginary parts we can get the series expansions for $\sin y$ and $\cos y$ valid for all real $y$.

It is interesting to note that I had the linked answer available with me for almost a year and yet it took your question to use the same technique to derive power series for circular functions. Things like these never cease to amaze me!
A: Note that Taylor's Inequality is valid for complex functions.

Let $$f:\;z\mapsto\lim_{n\to\infty}(1+{z\over n})^n.$$
One can prove that:


*

*$f$ is well-defined;

*$f$ is differentiable and $f'=f$ (thus $f \in \mathscr C^{\infty}$).


Then, choosing $M=3^r$ for example, one has
$$\forall r>0,\;\exists M>0,\;\forall z\in D(0,r),\;\forall n\in\Bbb N,\;|f^{(n)}(z)|\le M.$$
Hence, (using Taylor's Inequality), $f$ can be expressed by its Taylor series in $\Bbb C$, which is to say,
$$\forall z\in \Bbb C,\;f(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}.$$
A: Set, for $n\geq 1$, 
$$
(1+\dfrac{z}{n})^n:=\sum_{k=0}^\infty a(n,k)z^k
$$ 
(finite support, it is a polynomial). It can be checked easily (I can elaborate on request) that $a(n,k)$ is increasing in $n$ and then for all $j\geq 2$, one has $|a(j,k)-a(j-1,k)|=a(j,k)-a(j-1,k)$ 
Then, for all fixed $z\in \mathbb{C}$, the family $\Big((a(j,k)-a(j-1,k))z^k\Big)_{j\geq 2\atop k\geq 0}$, is absolutely summable as 
$$
\sum_{k\geq 0}|z^k|\sum_{j\geq 2}(a(j,k)-a(j-1,k))=\sum_{k\geq 0}|z|^k(\frac{1}{k!}-a(1,k))<+\infty\ .
$$ 
Now, as this family is absolutely (and then commutatively) summable, one has, summing it by columns
$$
\sum_{k\geq 0}z^k(\frac{1}{k!}-a(1,k))=e^z-(1+z)
$$ 
and, on the other hand, summing it by rows, 
$$
lim_{N\to +\infty}\sum_{j=2}^N\Big(\sum_{k\geq 0}z^k(a(j,k)-a(j-1,k))\Big)=lim_{N\to +\infty}\Big((1+\frac{z}{N})^N\Big)-(1+z)
$$ 
Thanks to @ParamanandSingh for having pointed the rôle of monotonicity (this is the golden mine of interaction). 
Note This result (and proof) holds for 
$$
lim_{n\to +\infty} (1+\frac{A}{n})^n=exp(A)
$$
where $A$ is a element in a complete associative commutative normed $\mathbb{R}$-algebra.
