weird properties of complex exponentials: $e^{i 2 \pi f(x)} = 1$

Question

$$ \begin{aligned} e^{i 2\pi f(x)} &= (e^{i\ 2\pi})^{f(x)} \\ e^{i 2\pi f(x)} &= (\cos {2\pi} + i\ \sin {2\pi})^{f(x)} \\ e^{i 2\pi f(x)} &= (1 + i 0)^{f(x)} \\ e^{i 2\pi f(x)} &= 1^{f(x)} \\ e^{i 2\pi f(x)} &= 1 \end{aligned} $$

True or False?

Answer 1

Long story short :

The property $(e^b)^c = e^{bc}$ holds for real numbers and in case of complex numbers you get such errors. (This is also the statement for any such exponent property). Wikipedia has an article-brunch analysis such false cases.

Also a side note, this question is broad. Even if it worked, without knowing what $f(x)$ is you can't say much. For example, if you allowed $f(x)$ to be equal to $\infty$ for some $x$ of its domain (Lebesgue Measure Theory cases), then $1^\infty$ is undefined.