Analyzing Euler's Identity 
From Wikipedia: "Euler's identity is a special case of Euler's formula
  from complex analysis, which states that for any real number x, 
$$ e^{ix} = cos(x) + isin(x) $$
where the inputs of the trigonometric functions sine and cosine are
  given in radians."

If $ e^{i\pi}=-1 $, then π is calculated in radians.
There appears to be an infinite number of solutions that yield -1 because x is calculated in radians, such that any value of $x$ that is a positive or negative odd multiple of π will yield -1
Examples:


*

*$$ e^{iπ} = -1 $$

*$$ e^{i(-π)} = -1 $$

*$$ e^{i3π} = -1 $$

*$$ e^{i(-3π)} = -1 $$

*$$ e^{i5π} = -1 $$

*$$ e^{i(-5π)} = -1 $$


Question: 
Being that $ cos(x) + isin(x) $ is identical to $ e^{ix} $
And cos(π) + isin(π) in degrees does not equal -1
Then is there a way to calculate an answer to $ e^{i\pi} $ without using degrees or radians? As you would with a simple problem like $$ 3^2=9 $$
 A: $$e^{x}=1+x+\frac{x^2}{2}+\frac{x^3}{6}\cdots$$ 
Based on the Taylor expansion rules
$$e^{ix}=1+ix+\frac{(ix)^2}{2}+\frac{(ix)^3}{6}...$$
You plug in $ix$ for $x$ and simplify all you notice
$$=1+ix-\frac{x^2}{2}-\frac{ix^3}{6}+\frac{x^4}{24}\cdots$$
which you can simplify into two [fairly recognizable] sums!
$$=(1-\frac{x^2}{2}+\frac{x^4}{24}\cdots)+i(x-\frac{x^3}{6}+\frac{x^5}{120}...)$$
By recogizable I mean sin and cos taylor expansions.
$$e^{ix}=\cos(x)+i\sin (x)$$
Plug in $\pi$ and you get $$e^{i(\pi)}=-1+i*0=-1$$ 
All you really have to know are Taylor Sums!
A: You can expand the Taylor polynomial for $e^{i\pi}$:
$$ e^{i\pi} = \sum_{k=0}^{\infty} \frac{(i\pi)^k}{k!} = 1 + i\pi - \frac{\pi^2}{2} \cdots$$
A: Yes! Just write $-1$ as you did. No radians at all in this representation of $e^{i\pi}$. The only thing is that there is more than one way to write this number, by using radians. In fact, any complex number, except the zero, can be written in infinitely many ways with the form $\cos(x)+i\sin(x)$ and in only one with the form $a+ib$.  
