Euler's formula, is this true? *I've changed this question as below.
Let me have a function such as $ f(k) = \exp(j 2 \pi k ) $,  where $k$ is real value.
Using Euler's formula, we can write $f(k)$ as below,
$$ f(k) = \exp(j  2 \pi  k ) = \cos(2\pi k)+j \sin(2\pi k).$$
If $k$ is integer, this always goes to 1.
Up to here, nothing is weird and makes sense.
However, if the equation goes to
$$f(k) = \exp(j2\pi k)=e^{j2\pi k}=(e^{j2\pi})^k = (\cos(2\pi)+j\sin(2\pi))^k=1^k=1$$
What I want to know is this above equation makes sense or not.
Thank you in advance.
 A: The issue that you are having is that $z^k$ can be a multivalued function, dependent on $k$. As a basic, example, we all know that $(2)^2 = 4 = (-2)^2$. So we can just as well define $4^{1/2}$ as $+2$ or $-2$.
By your argument, we can have (the clearly erroneous result)
$$ -1 = e^{\pi i} = e^{2 \pi i \cdot {1 \over 2}} = (e^{2 \pi i})^{1/2} = 1^{1/2} = 1.$$
But, as with my example above, but using $(1)^2 = 1 = (-1)^2$, we see that we can define $1^{1/2}$ as either $+1$ or $-1$. The technical complex analysis term is a 'branch cut' - using one of these allows you to define $z^k$ uniquely.
I've answered a question before on a similar topic - Why do I get two different results for the reciprocal of i?.
Hopefully this answer has been helpful to you! If it has, then please remember to upvote and/or accept! Hope you understand now! :)
A: Following on from Yves Daoust:
$$\left(z^x\right)^y$$
$$=\left(e^{\log \left(z^x\right)}\right)^y=\left(e^{x\log \left(z\right)}\right)^y$$
$$=e^{\log \left(\left(e^{x\log \left(z\right)}\right)^y\right)}=e^{y\log \left(e^{x\log \left(z\right)}\right)}$$
$$= e^{y\left(x\log \left(z\right)+i\left(2nπ\right)\right)}$$
$$=z^{xy} ⋅ e^{i\left(2nyπ\right)}$$
A: For non-integer exponents, exponentiation $x^y$ is only well defined (denotes a single value) when $x$ is a positive real number; on the other hand $y$ can be any complex number. (And for most purposes one can do with just using $x=e=\exp(1)$, but that is a separate matter). Assuming that, the rule $x^{y+z}=x^yx^z$ is always valid. However the rule for multiplication in the exponent has a restriction that is rarely stated explicitly:
$$
  x^{yz}=(x^y)^z \qquad
  \text{provided that $x>0$ is real} \textbf{ and that } \text{$y$ is real.}
$$
Note that it does not suffice that the expressions on both sides are defined, in other words that $x$ and $x^y$ are positive real numbers: both sides can be well defined value, yet different. The example in the question illustrates this (for any non-integer value of$~k$).
See this answer to a similar question for more details, and a proof of the rule stated above.
A: No, in complex numbers you may not assume that $(e^a)^b=e^{(ab)}$. You just found a counter-example.
