calculus - 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 $$ 3^2=9 $$

Answer

$$e^{x}=1+x+\frac{x^2}{2}+\frac{x^3}{6}\cdots$$
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!