Why isn't $\sqrt[i\pi]{-1} = e$? My reasoning is through Euler's identity 
$$e^{i\pi} = -1$$
$$\sqrt[i\pi]{e^{i\pi}} = \sqrt[i\pi]{-1}$$
$$e = \sqrt[i\pi]{-1}$$
but Wolfram Alpha says otherwise,  $\sqrt[i\pi]{-1} = -\pi$
So the way I see it, there are three options:    


*

*Wolfram Alpha is wrong    

*$-\pi = e$    

*I am wrong


I'm pretty sure it's the latter. But why?
 A: If you look at the input field of the wolframalpha link, you would see that wolframalpha interpreted it as $(i\pi)\sqrt{-1}$.
And as to $\sqrt[i\pi]{-1}$, we are trying to solve $x^{i\pi}=-1$, i.e. $i\pi\ln x = i\pi+2ni\pi$, i.e. $\ln x = 2n+1$, i.e. $x=e^{2n+1}$, so any number of the form $e^{2n+1}$ is an $i\pi$-th root of $-1$.
A: *

*"Wolfram Alpha is wrong"


It is.  It read this as $i\pi * \sqrt {-1} = i\pi * i = - \pi$.
Typing in $(-1)^{\frac 1{i\pi}}$ (I wasn't sure how to type in $n-th$ roots) Wolfram Alpha gave $e$.


*

*"$-\pi = e$"


It doesn't.  But you should be careful about multiple values.  One can make a similar mistake that
$e^{2\pi i} = 1$ so
$\sqrt[2\pi i]{e^{2\pi i}}= \sqrt[2\pi i]{1} = \sqrt[2\pi i]{1^{2\pi i}}$
$e = 1$.
Which is obviously false.  
But in actuality, $e^{k*2\pi i}$ for $k \in \mathbb Z$ so $\sqrt[2\pi i]{e^{k*2\pi i}}= \sqrt[2\pi i]{1}$ actually means $1 \in  \{e^k; k \in \mathbb Z\}$.  Which it is.  $1 = e^0$.
In your case $e$ is not the only possible value for $\sqrt[\pi i]{e^{\pi i}} = \sqrt[\pi i]{e^{\pi i+ 2k\pi i}}$ but $e^{1 + 2k}$ will be too.


*

*I am wrong.


Well,... not really.  But you have to watch out for multivalues and realize the answers you get will be congruence and not exact single values.  So when you do get something like $-\pi  = e$ one should wonder if there is some congruence between $-\pi$ and $e$.  But there isn't in this case.
===
Actually the entire concept of the $n$-th root $\sqrt[n]{}$ is a concept that doesn't really work well with complex numbers and is best avoided all together.
If one needs the "$z$th root" of something.  i.e. $w^z = v;$ solve for $w$. one should solve $z \ln w = \ln v$ bearing in mind that $\ln v$ is multivalued congruences modulo $2\pi i$.
