I was experimenting with Euler's Identity. If $e^{((\pi*i)/2)}$ is $i$, couldn't you recursively plug in $i$ into the expression. For example: $$e^{((\pi/2)*e^{((\pi/2)*e^{((\pi/2)*e^{((\pi/2)*e^{((\pi/2)*e^{((\pi/2)*e^{((\pi/2)i)))))))}}}}}}}. $$
That should be equal to $i$. So you could plug that whole thing in for the $i$ in the equation, giving a new equation equaling $i$, so that new equation can be plugged into itself. If you do this an infinite number of times, you wind up with an infinitely large power tower which uses only positive numbers, and equals $i$. The farther you go to this expression, the closer it should get to $i$. Logically it can never equal $i$ when you apply the recursion a finite number of times, but it can go to $0$, which is the closest it can get. However, when you plug this in, it quickly zooms off towards infinity, not $0$. Could you help me resolve this?
Your "infinite tower" indeed does not equal $i$, but rather shoots off to infinity. The error is your sentence
specifically, the part in bold.
Basically, your problem is that the evaluation of an infinite expression has to be done very carefully.
Here's a much simpler version of your paradox:
Clearly $1\times 0=0$.
So we can plug in "$1\times 0$" in place of "$0$", above, to get $1\times 1\times 0=0$.
We can iterated this indefinitely, and the "limit expression" is $1\times 1\times 1\times 1 \times\dots $
. . . but this clearly isn't equal to $0$!
What's going on, basically, is that:
The sequence of expressions $1\times 0, 1\times 1\times 0, 1\times 1\times 1\times 0, \dots$ does approach (in an appropriate sense) the infinite expression $1\times 1\times 1\times \dots$.
However, the evaluation operation - think of this roughly as a function $ev$ from $\{$expressions$\}$ to $\{$numbers$\}$ - is not continuous in the appropriate sense: that is, $$\mbox{$\lim_{i\rightarrow\infty} L_i=L\quad$ does not imply $\quad \lim_{i\rightarrow\infty} (ev(L_i))=ev(L)$.}$$
This, in turn, is just a recasting of the standard example from calculus: let $f_i(x)=1$ if $x>i$, and $0$ otherwise; then $\lim_{i\rightarrow\infty}\lim_{x\rightarrow\infty}f_i(x)=1$, but $\lim_{x\rightarrow\infty}\lim_{i\rightarrow\infty}f_i(x)=0$. Do you see why?
(Note that I'm sweeping a whole ton under the rug in terms of what "$ev$" is exactly - my point is just that, merely because we have some way of evaluating some infinite expressions, doesn't mean that that evaluation method is "nice" in any intuitive sense.)
A similar issue is at play in the popular joke argument that $\pi=4$ (see https://qntm.org/trollpi); what's going on is that, depending on how we topologize the set of all "reasonable" curves in the plane,
EITHER the "length" function is not continuous,
OR the sequence of jagged approximations doesn't actually approach the circle.
Depending on what you're trying to do, one or the other option may be better; but you can't avoid both simultaneously!