Likely fake proof of the irrationality of a rational multiple of pi/pi I would like to know what has gone wrong in this 'proof'. 
Suppose that $$k\frac 1\pi\pi=\frac ab\operatorname{,where}a,b\in \mathbb Z\operatorname {and}k\in \mathbb Q.$$
Then, we multiply both sides by $i$ and raise  $e$ to the power of both sides. We find that $$e^{i(k\frac 1\pi)\pi}=e^{i\frac{2a}{2b}}$$ This implies that $$e^{i(2bk\frac 1\pi)\pi}=e^{i2a}.$$ Using laws of expontents, we rewrite this as $$\begin{pmatrix}e^{i2b\pi}\end{pmatrix}^{\frac k\pi}=e^{i2a}.$$ Since b is an integer, the left hand side is equal to $1$. We take the natural log of both sides using the definition of the natural log in the complex domain.Thus,$$0=i(2a+2n\pi)\operatorname{for some integer }n.$$ Since the imaginary part of $0$ is $0$, $$a+n\pi=0.$$ Rearranging the equation for $\pi,$ we find that $\pi=-\frac an$, which is a contradiction as $\pi$ is irrational.
This is very odd because $k\frac 1\pi=k$ and $k$ is assumed to be a rational number. I have no idea what I have proven despite trying being careful, particularly when using complex numbers. Could someone point out an error?   
 A: The main problem is that some of the rules for working with powers from real numbers are no longer true with complex numbers.
While for 2 complex number $z_1,z_2$ the rule $e^{z_1+z_2}=e^{z_1}e^{z_2}$ is still correct, the version dealing with multiplied exponents isn't, generally we have
$$e^{z_1z_2} \color{red}\neq  \left(e^{z_1}\right)^{z_2}.$$
A simple example is $z_1=2\pi i, z_2=\frac12$, where the left hand side is $e^{\pi i}=-1$ while the right hand side is $1^{\frac12}=1$.
The reason why that is so is that even defining what the right hand side means is non-trivial in the general case. I refer you to the  Wikipedia entry for Complex exponentiation for further information.
In your proof, where you go wrong is in rewriting $e^{i(2bk\frac1{\pi})\pi}$ as $\left(e^{i2b\pi}\right)^{\frac{k}\pi}$, because that is using exactly the above incorrect formula for multiplied exponents.
Note that you did a similar step before, when you went from
$$e^{i(k\frac 1\pi)\pi}=e^{i\frac{2a}{2b}}$$
to
$$e^{i(2bk\frac 1\pi)\pi}=e^{i2a}.$$
This is not incorrect, though I guess just "by accident". You muliplied both exponenets by the integer $2b$, which works, for any complex $z$ and any integer $n$ we have
$$e^{nz}=\left(e^z\right)^n$$
That's a consequence of $n$ being an integer and using the valid rule about adding exponents:
$$e^{nz}=e^{z+z+\ldots+z}=e^ze^z\ldots e^z=(e^z)^n,$$
if $n$ is positive and for negative integer $n$ it then follows from $e^{-z}=\frac1{e^z}$.
To summarize, dealing with complex exponentiation can be hard because internalized rules that were valid for real numbers no longer apply, or only in special cases. Any time you see a potentially non-real complex number raised to a non-integer power is cause for thinking really hard what that means in the given context. Same goes for a real number raised to a complex power. 
A: There is no error there, except the error of change of definition
You know that $e^{\pi*i} = -1$
Which implies that $e^{2*\pi*i} = 1$
But we know also know that $e^0 = 1$
Does this mean that $2*\pi*i = 0$?
We know that $\sqrt(4) = ±2$
Does this mean than $+2 = -2$?
Mathematics is biased, so when twisting definition we should be careful
