The proof of the formula $z = |z|e^{i \phi}$ I would like to proof that:
$$z = |z|\big(\cos(\phi) + i \sin(\phi)\big)$$
of course the second part can be shown using Euler's formula. That's why I would like to prove that:
$$z = |z|e^{i \phi}$$
I can use only the series definition of exp and some basic properties like $e^{a+b} = e^ae^b$.
I think it must be quite trivial but I really don't know how to do it properly.  
I think that my problem is a bit different to the one in the suggested  post. I do know how to prove that $e^{i\phi} = \cos \phi + i\sin \phi$ but I do not know how to prove that 
$$\forall_{z \in \mathbb{C}} z = |z|\exp(ix)$$ for some $x \in \mathbb{R}$. 
 A: I don't know if this is what you want, but you can alwyas prove Euler's formula:\begin{align}e^{x+yi}&=e^xe^{yi}\\&=e^x\left(1+yi+\frac{(yi)^2}{2!}+\frac{(yi)^3}{3!}+\cdots\right)\\&=e^x\left(\left(1-\frac{y^2}{2!}+\frac{y^4}{4!}-\cdots\right)+\left(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots\right)i\right)\\&=e^x\bigl(\cos(y)+\sin(y)i\bigr).\end{align}
A: Every $z\in\mathbb{C}$ can be written as $$z=x+iy$$
with $x,y\in\mathbb{R}$. By definition $|z|:=\sqrt{x^2+y^2}$. If $\theta$ is the angle the vector $(x,y)$ makes with the horizontal axis (the real axis) in the positive direction then $x=|z|\cos\theta$ and $y=|z|\sin\theta$. Therefore
$$z=x+iy=|z|\cos\theta+i|z|\sin\theta=|z|(\cos\theta+i\sin\theta)=|z|e^{i\theta}$$
where the last equality follows from Euler's identity
$$e^{i\theta}=\cos\theta+i\sin\theta$$
A: If
$z \ne 0, \tag 1$
then
$\left \vert \dfrac{z}{\vert z \vert} \right \vert = \dfrac{\vert z \vert}{\vert z \vert} = 1; \tag 2$
thus
$\dfrac{z}{\vert z \vert} \in S^1, \tag 3$
the unit circle in $\Bbb C$; therefore, we may always find some $\phi \in \Bbb R$ with
$\dfrac{z}{\vert z \vert} = \cos \phi + i \sin \phi; \tag 4$
it remains to show that
$\cos \phi + i \sin \phi = e^{i \phi}; \tag 5$
but this may easily be done by expanding $e^{i \phi}$ in a power series
$e^{i\phi} = \displaystyle \sum_0^\infty \dfrac{(i\phi)^n}{n!}, \tag 6$
and separating out the real and imaginary parts, as is shown in this wikipedia article, as well as by Jose Carlos Santos in his answer.  Then (4) becomes
$\dfrac{z}{\vert z \vert} = \cos \phi + i \sin \phi = e^{i\phi}, \tag 7$
or
$z = \vert z \vert e^{i\phi}. \tag 8$
Note Added in Edit, Tuesday 3 April 2018 10:34 AM PST:  This in response to the comment to this answer made by our OP Hendrra.  The easiest way to see that 
$z \in S^1 \Longrightarrow \exists \phi \in \Bbb R, \; z = \cos \phi + i \sin \phi, \tag 9$
is via simple geometry and trigonometry.  Since $z$ is a point on the unit circle, there is a radial line segment 'twixt the origin $O$ and $z$, $\overline{Oz}$; the length of this segment is $1$, since $S^1$ is the "unit circle".  Then let $\phi$ be the angle 'twixt the positive $x$-axis and the segment $\overline{Oz}$; the $x$-coordinate of the point $z$ is then the real part of $z$ considered as a complex number:
$\Re(z) = \vert \overline{Oz} \vert \cos \phi = \cos \phi, \tag{10}$
since $\vert \overline{Oz} \vert = 1$; likewise the $y$-coordinate is
$\Im(z) = \sin \phi; \tag{11}$
thus
$z = \Re(z) + i\Im(z) = \cos \phi + i \sin \phi. \tag{12}$
End of Note.
