Does $\exp(2ir\pi)$ equal $1$? What's wrong?

Question

Since $e^{ix}=\cos x+i\sin x$,
thus $e^{2\pi i}=\cos2\pi+i\sin2\pi=1$

Now I take arbitrary real number $r$
then $e^{i2\pi r}=(e^{i2\pi})^r=1^r=1?$

But this cannot be true since $\cos2\pi r+i\sin2\pi r$ isn't necessarily $1$ for arbitrary $r$. Where did it go wrong?

Answer 1

The answer is simple:

$$(a^b)^c = a^{bc}$$

is an equality which only holds for real numbers with $a>0$ and does not necesarily hold on complex numbers.