Is it okay to raise to the power of $n$ (where $n\in\mathbb{R}$) on both sides in Euler's Identity?

Question

So I was playing around with the identity and did this: $$e^{i\pi}=-1$$ $$e^{(2\pi)i}=(-1)^{2}=1$$ $$e^{(2n\pi)i}=1^{n}=1\ \ \ \ \ (n\in \mathbb{R})$$ $$\Rightarrow\ e^{(2n\pi)i}=e^{(2\pi)i}\ \ (\forall n\in\mathbb{R})$$

Then, $$2n=2$$ So,$$n=1$$

Which is quite contradicting as any value of n should satisfy the condition. Which means I have made a mistake somewhere. Can anyone tell me where am I going wrong? Help would be appreciated! Thanks in advance.

Answer 1

$$1^m =1^n $$ does not imply $$m=n$$

The positive real number base $b$ should be different from 1 to have $$b^m=b^n $$implies $$m=n$$

Answer 2

No. But the ever useful De Moivre's formula falls out by doing $e^{ni\theta}=\cos(n\theta)+i\sin(n\theta)$.