Show that $e^{\frac{\pi}{2}i }= i$ - Problem to understand a certain convergence We have defined $e^{ix}=\cos x+ i \sin x$ where $x\in\mathbb{R}$.
I also have proved earlier that $\cos(\frac{\pi}{2})=0$ and know that $\sin(0)=0$
Now I want to prove that 
$e^{\frac{\pi}{2}i }= i$
I know that 
$|e^{xi }|= 1,\forall_{x\in\mathbb{R}}$
Therefore
$1=|e^{\frac{\pi}{2}i}|=|\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})|=|\sin(\frac{\pi}{2})|$

And now the part that I don't understand
Because $\sin'(x)=\cos(x)>0,\forall_{x\in[0,\frac{\pi}{2})}$
$\Longrightarrow\sin(\frac{\pi}{2})=1$ 

$\Longrightarrow e^{\frac{\pi}{2}i }= i$
Please explain me which Theorem we use here. And how this Situation can be generalised, so if I Encounter something similiar I directly know what to do
Edit:
Can it be shown constructively, i.e. with the mean-value-theorem for example?
 A: You know that $ \sin(x) \in \mathbb{R}$, so that 
$$\left\vert \sin \left( \frac{\pi}{2} \right) \right\vert=1 \Rightarrow \sin \left( \frac{\pi}{2} \right)= \pm1$$
If $\sin'(x)=\cos(x) >0$ for all $x \in [0,\frac{\pi}{2})$, then this means that the sinus function is strictly growing on $x \in [0,\frac{\pi}{2})$. With $\sin(0)=0$, with $\sin$ being an continuous function, then the only possibility is $$\sin \left( \frac{\pi}{2} \right)=+1$$
Hence $$e^{i\frac{\pi}{2}} = i$$
Edit If you want to proove that $\sin'(x)=\cos(x)$, so note that $$\frac{d}{dx}\left( e^{ix} \right) = ie^{ix} = i\cos(x) - sin(x)$$
Implies that 
$$ \cos'(x) + i\sin'(x) = -\sin(x) + i\cos(x)$$
Prooving that $\sin'(x)=\cos(x)$
A: Essentially we need only to prove $\sin\frac\pi2>0$. The answer will use the mean value theorem (MVT) as asked in question.
The definition of $e^z,\sin z, \cos z$ implies that the functions are continuous and differentiable. Particularly:
$$
\forall z\in\mathbb C: \sin'z=\cos z.\tag1
$$
As $\frac\pi2$ is by definition (see discussion in comments) the least positive root of $\cos x$ and $\cos(0)=1>0$ we have from the continuity of the function:
$$
\forall x\in\left(0,\frac\pi2\right): \cos x>0.
$$
Together with (1) this implies:
$$
\forall x\in\left(0,\frac\pi2\right): \sin' x>0.\tag2
$$
Now by MVT there exists such a point $0<c<\frac\pi2$ that
$$
\sin'(c)=\frac{\sin\frac\pi2-\sin0}{\frac\pi2-0}
\implies \sin\frac\pi2=\frac\pi2\sin'(c).
$$
As both $\sin'(c)$ and $\frac\pi2$ are positive, we obtain $$\sin\frac\pi2>0.$$
