How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$? Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
Proof:
Consider the function $f(t) = e^{-it}(\cos t + i \sin t)$ for $t \in \mathbb{R}$. By the product rule 
\begin{eqnarray}
f^{\prime}(t) = e^{-i t}(i \cos t - \sin t) - i e^{-i t}(\cos t + i \sin t) = 0
\end{eqnarray}
identically for all $t \in \mathbb{R}$. Hence, $f$ is constant everywhere. Since $f(0) = 1$, it follows that $f(t) = 1$ identically. Therefore, $e^{it} = \cos t + i \sin t$ for all $t \in \mathbb{R}$, as claimed.
An easy way out to answer the question is when we try to extend the definition of exponential to complex plane in a "nice" way (read "nice" as holomorphic) we then end up with this definition. And as you are probably aware there is only at most one such extension. So if we want to define our exponential in the complex plane such that the exponential function is holomorphic, and it matches with the conventional exponential function on the real line, we end up with $e^{it} = \cos(t) + i \sin(t)$. 

All the other answers I think are circular in some sense. 

For instance, how do we know that we can expand $e^{i x}$ as a power series. All we know is that we can expand $e^{x}$ as a power series when $x$ is real. We do not know apriori that the expansion can be carried forward if $x$ happens to be complex.