To completely understand why $\cos\theta+i\sin\theta=e^{i\theta}$, you need to understand Maclaurin series. Using the Maclaurin series, we extend the domain of the exponential function to the field of complex numbers $\mathbb C$. But, if you don't know Maclaurin series yet, you can still gain sufficient understanding to feel comfortable using this formula.
For the time being, it is useful to view this formula just as a "convention of notation" - a compact way of writing $\cos\theta+i\sin\theta$.
Why do all the formulas work?
This fact is due to the following formulas:
\begin{eqnarray}
&(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2) = (\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))& \tag{1}\\
&(\cos\theta+i\sin\theta)^n = (\cos n\theta + i\sin n\theta)& \tag{2}
\end{eqnarray}
The first formula can be easily proven by just cross-multiplying the LHS and using the additive trigonometric formulas for cos and sin.
One can see that the argument of the result is the sum of the arguments of the two inputs, just like when we multiply two complex numbers in Euler form.
The second formula can be easily proven for $n \in\mathbb N$ using mathematical induction, although it is valid for any $n \in\mathbb R$. The formula is called Moivre's formula.
Even if you do not know how to prove these, the proofs are easy to find on the internet.
As for why division works, use that
$$\frac{e^{i\theta_1}}{e^{i\theta_2}}=e^{i(\theta_1-\theta_2)}$$
is equivalent with
$$e^{i\theta_2}e^{i(\theta_1-\theta_2)}=e^{i\theta_1}$$
Changing this into trigonometric form and using formula (1), one can see why this equality holds.
