What's wrong with this proof $1=i^2=-1$ I just thinking what $i^{i}$ should be, arrived at a quiet awkward thing. 
So this was that awkward thinking :
Let $ i^{i} = a$
$(i^{i})^{2i} = a^{2i}$
$i^{-2}=a^{2i}$
$-1=a^{2i}$
Now if we take power of $\frac{1}{2i}$ both sides 
$(-1)^{\frac{1}{2i}}=a$
$((-1)^{\frac{1}{2}}) ^{\frac{1}{i}}=a$
$i^{\frac{1}{i}} =a$
$i^{\frac{1}{i}} =i^{i}$
Now if we compare the power (since the base is same) 
$\frac 1i = i$
$i^2=1$
So,  $-1=i^2=1$. 
I am quite new to imaginary numbers,  So please show my mistake elaborately. 
P. S. Thanks in advance. 
 A: There are two illegal steps you've made:


*

*When you use $x^{ab} = (x^a)^b.$ This type of arithmetic is also not generally valid for complex numbers. A counterexample is when $a = 2$ and $b=1/2$ and $x=-i.$ Then $(-i)^{2\cdot 1/2} = (-i)^1=-i,$ but $((-i)^2)^{1/2} = (-1)^{1/2} = i.$ Likewise, if $x^a = y^a$ we cannot take both sides to the $1/a$ power to give $x=y.$ For instance $(-2)^2 = (2)^2$ but $-2\ne 2.$ 

*When you conclude the exponents are equal cause the bases are equal, i.e. conclude from $a^b = a^c$ that $b=c.$ A counterexample to this is that $e^{x+2\pi i} = e^x$ for any $x$ but the exponents are not equal. 
The difficulty for 1 is that roots are generally not unique, so when we implicitly pick a 'principal' root during arithmetic it might not be the right one. For the second it's that the exponential function is not one-to-one when the domain is extended to the complex numbers, so we can't invert it by taking logs like the real case. If $e^a = e^b,$ we know $a$ and $b$ are different by an integer multiple of $2\pi i,$ not that they are equal.
A: I'll give you another one. 
$$1=\sqrt{1}=\sqrt{(-1)^2}=\sqrt{-1}\sqrt{-1}=i^2=-1.$$
For positive reals $a$ and $b$ we have that $\sqrt{ab}=\sqrt{a}\sqrt{b}$. But who says this still holds for other numbers?
