Is $\infty$ the solution to ${x}^{{x}^{{x}^{{x}^{x\dots}}}} = i$? I started off trying to solve the problem ${x}^{{x}^{{x}^{{x}^{x\dots}}}} = i$ where $i$ is the imaginary unit, and with an infinite amount of $x$'s. I then substituted the $x$'s in the exponents as $i$, and got $${x}^{i} = i$$    After solving this, one of the primary solutions I got was $x = \mathrm{e}^{\pi/2}$. but substituting this into the original equation, we get $\infty = i$ since the infinite tetration of $\mathrm{e}^{\pi/2}$ is $\infty$. So is infinity really equal to the imaginary unit, or did I make a mistake?
Edit: I am aware that for something like the equation $\mathrm{x}^{y} = y$, it can only be solved as long as y is equal to or less than $e$ and equal to or more than $
{1 \over e}$, but since i is an imaginary number, i am completely stumped on whether or not this should work.
 A: No! You've made a rather big error in your logic. Your algebraic manipulations prove an implication, not an equivalence - so $x^i=i$ has solutions that are not solutions to your original equation. 
In particular, to hide any infinite weirdness, let's just write $f(x)=x^{x^{x^{\ldots}}}$ and not worry too much about how to define this*, except that we want $x^{f(x)}=f(x)$. You're trying to solve
$$f(x)=i$$
 So, the equation $f(x)=i$ then implies, by substituting $x^{f(x)}=f(x)$, that
$$x^{f(x)}=i$$
which then, substituting $f(x)=i$ yields
$$x^i=i.$$
This means that every solution to $f(x)=i$ is a solution to $x^i=i$. It does not mean that every solution to $x^i=i$ is a solution to $f(x)=i$. In particular, just because $e^{\pi/2}$ satisfies $x^i=i$ that doesn't mean it satisfies $f(x)=i$ - irrespective of whether we think $f(x)=\infty$ or not.
This is not a special problem with the question - one frequently encounters this issue. For instance, if I was trying to solve
$$x+1=2$$
it is perfectly correct to square both sides and write
$$x^2+2x+1=4$$
but one has to understand that the second equation has two solutions, which are $x=1$ and $x=-3$, and this doesn't imply that they solve the first equation. In general, algebraic manipulations are only valid in one direction, unless they can be undone by another manipulation (e.g. multiplying an equation by $2$ can be undone by dividing by $2$, but multiplying an equation by $x$ can't be undone if $x=0$ - which might lead to a new solution).

*Okay, if we were being a bit more careful, we might try to define
$$f(x)=\lim_{n\rightarrow\infty}x^{x^{\ldots^x}}$$
where there are $n$ copies of $x$ in the tower. There some problems here - namely that this sequence might not converge, even if we had some meaningful** notion of $\infty$. We could prove that, wherever it does converge, we have the relation
$$x^{f(x)}=f(x)$$
so a solution to $f(x)=i$ does truly imply $x^i=i$.
This sort of definition is pretty fragile though - there are cases that look okay, but act really badly. For instance, maybe if we were instead trying to solve
$$x=1+\frac{1}x$$
we would try substituting and write
$$x=1+\frac{1}x=1+\frac{1}{1+\frac{1}x}=1+\frac{1}{1+\frac{1}{1+\frac{1}x}}$$
which is a consequence of the the last equation (though, again, will create extraneous solutions if we just set the first and last equation equal). We'd be really tempted to write
$$x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$$
and to make the $x$ entirely vanish... but this doesn't work for subtle reasons - in particular, if we start with $1$ and make this substitution, the sequence $1,1+\frac{1}1,1+\frac{1}{1+\frac{1}1},\ldots$ actually does converge (to $\frac{1+\sqrt{5}}2$, which is a solution to $x=1+\frac{1}x$), but if we start with $\frac{1-\sqrt{5}}2$ - which is the other solution to $x=1+\frac{1}x$ - when we start substituting, we get a different answer - which tells us that our vanishing trick with the $x$ was not legal and that we would instead have to write out an honest limit. This issue arises pretty much universally when you have an "infinite" expression - you are basically doomed to say exactly what you mean, and then you might find out that the behavior of the whole object depends on what happened in the very beginning - which is, inconveniently, the part that you entirely conceal with "$\ldots$".
**While $\lim_{n\rightarrow\infty}x^{x^{\ldots^x}}$ goes to $\infty$ when $x$ is a real number greater than $1$, this is a fairly delicate statement that does not carry over well into the context of complex numbers mostly because definitions of such limits rely on the ordering of the real numbers. The complex numbers are not ordered, so the usual definitions simply don't apply. What's worse is that, in general, if a mathematician is talking about $\infty$ in the complex numbers, they far more commonly mean a concept that is completely different than what they meant with the symbol in the real numbers.
A: Short answer: no.
Longer answer. $\infty$ is not a number. Expressions that seem to involve infinitely many operations are hard to define. You can't naively assume the usual rules of arithmetic work. For example
$$
2 \times \infty = \infty + \infty = \infty = 1 \times \infty
$$
does not imply 
$$2 = 1 .
$$
Whenever you reach contradictions like the one in the question you can probably trace it back to some unfounded assumption behind a formal argument.
A: I don't want to spoil the value of @Milo Brandt's really nice answer. Just one aspect which I think is often missed in questions like this about infinite exponential towers.          
Actually, the infinite expansion of the basic equation $x^i=i$ is 
$$  x^i = i \qquad  x^{x^i}=i \qquad  x^{x^{x^i}}=i \qquad \cdots
$$
and this is only true when always $i$ is on the top of the expression, defining an initial value (the fixpoint) of the "unbounded-towards-leftdown" tower
$$ \tiny{.\cdot}^{ \Tiny{ x}^{ \small{x}^{\large{x}^{\Large {i}}}}} \large {=i} $$
This notation is then even correct with an infinite set $\Lambda$ of different fixpoints $\lambda_k \in \Lambda$ writing always
$$ \tiny{.\cdot}^{ \Tiny{ x}^{ \small{x}^{\large{x}^{\Large {\lambda_{\small k}}}}}} \large {=\lambda_k} $$
For instance, using your value  $x=e^{\pi/2}$ in the same way but with approximate $\lambda_2= 1.02132 - 4.86835 î $  you would have
$$ \tiny{.\cdot}^{ \Tiny{ x}^{ \small{x}^{\large{x}^{\Large {1.02132 - 4.86835 i}}}}} \large {=1.02132 - 4.86835 i} $$
Or with approximate $\lambda_3=1.39951 - 8.90071 i$
$$ \tiny{.\cdot}^{ \Tiny{ x}^{ \small{x}^{\large{x}^{\Large {1.39951 - 8.90071 i}}}}} \large {=1.39951 - 8.90071 i} $$

Thus your own writing with the infinite tower of $x$ with $x$ itself on top (which is however much common here in MSE) is only meaningful as long there is    


*

*an $x$ from the interval $1/e^e \le x \le e^{1/e}$ for real $x$   

*or $x$ is in the Shell-Thron-region for complex $x$. 

A: If your infinite tower is defined, $x^i=i$. But no $x$ solving this equation works, so the tower is undefined.
A: If $x^i=i$, then $x=i^{1/i}$.
