Question about $i$ Was just fooling around with some identities and made this monster:
Step (1) $i^{4} = 1$
Step (2) $ \sqrt[4]{i^{4}} = \sqrt[4]{1} $
Step (3) $i = 1$
Step (4) $\sqrt{-1} = 1$
What did I do wrong here?
Update: Thank you! All of you. For your time and understanding. I've been fooling around with some identities (trying to see how many things I can get to equal each other by setting them equal to 1).
Also, excuse my informal language. I never was one to..."adopt" math-speak. I'm just a guy kinda interested in math.
I think I'll stay away from $i$ for a while.
 A: Complex numbers are a red herring; you can still see this with ordinary real numbers: $(-1)^4 = 1^4$.
The general point is that $x^4 = y^4$ does not imply $x = y$. Recall that every positive number has two square roots: if you are told that $x^2 = 4$, then there are two different solutions for $x$: $x=2$ and $y=2$. Similarly, if $x^2 = y^2$, then you could have $x=y$ or you could have $x = -y$.
This extends in full generality to complex numbers. e.g. if $\omega = \frac{1}{2}(-1 + \mathbf{i} \sqrt{3})$, which is a cube root of $1$, then you have the theorem:

Theorem If $x^3 = y^3$, then either $x = y$, $x = \omega y$, or $x = \omega^2 y$.

And similarly, 

Theorem If $x^4 = y^4$, then either $x = y$, $x = \mathbf{i} y$, $x = -y$, or $x = -\mathbf{i} y$.

and so on for higher powers.
If you're interested in further reading, some keywords are "multivalued function" and "complex exponential".
And maybe "principal value" as well, which is analogous to our habit of always selecting the positive square root when working with positive real numbers, although reading on the topic is likely to explain why it doesn't work nearly as well in this general case.
A: Your mistake is in step 2. As David K pointed out in the comments, $\sqrt[4]1$ is not strictly $1$, it could be $-1$ or $\pm i$. It is the same with square roots- $$(-1)^2=1$$ Then take the square root of both sides-$$\sqrt{(-1)^2}=1\\ \Longrightarrow-1=1$$
A: To add to the answers already given in the comments above, when one speaks of "taking the $n$th root", the operation which is denoted $\sqrt[n]{x}$, it is usually in the sense of the arithmetic root, which is only defined when $x$ is a real number and $x\ge0$, as the unique non-negative real number $y$ such that $y^n=x$. In general, however, as noted above, many other numbers (negative and complex) will satisfy this equation, hence why we specify only one value as the canonical root. Therefore, it is easy to see from the definition that $\sqrt[4]{i}$ does not really make any sense at all; $i$ is not a real number. 
A: The wrong step is the third. I assume that you meant $\sqrt[4]{i^4}$ in the second step.
The third step is wrog because $\sqrt[4]{i^4}$ is $1$, not $i$. The symbol $\sqrt[4]x$ is the real, positive number $y$ such that $y^4=x$, so in an expression of the form $\sqrt[n]{z^n}$, the index and the exponent  can be cancelled only if $z$ is a positive, real number, or zero.
A: It's excellent that you've stumbled upon this now! It's something you may see in the future: 
Technically, the root of a number isn't well defined. That is, the root "function" gives multiple values as its output. For the sake of simplicity, we usually restrict the output. We call all the roots of $1$ the roots of unity, for example. (It may be worth noting that, in mathematics, whenever you see the term unity, it usually refers to something that's equivalent to the number $1$).
I've stolen this definition directly from Wikipedia:

In mathematics, a root of unity [...] is any complex number that gives 1 when raised to some integer power, $n$.

There's something called Euler's Formula: 
$$e^{i \theta} = \cos{\theta} + i\sin{\theta} \implies re^{i\theta} = r(\cos{\theta} + i \sin{\theta})$$ Which we use extensively in complex analysis. 
There is another thing to know: We can write any complex number as $~re^{i \theta}~$, and thus write any complex number as $~r(\cos{\theta} + i\sin{\theta})~~^{[1]}$, by Euler's Formula above. Let's say $z$ is some arbitrary complex number. We can thus write it as $re^{i\theta}$. Now, take the $4^{th}$ root of this number: 
$$\sqrt[4]{z} ~=~ z^{\frac{1}{4}} ~=~ (re^{i\theta})^{\frac{1}{4}} ~=~ re^{i \frac{\theta}{4}} ~=~ r\left(\cos{\frac{\theta}{4}} + i\sin{\frac{\theta}{4}} \right)$$
Remember from trigonometry: $\cos{\theta} = \cos{(\theta + 2\pi n)}$ where $n = 0, 1, 2, 3, ...$ This works the same way for $\sin{\theta}$. That is, you can add any multiple of $2\pi$ to $\sin$ or $\cos$ and the values don't actually change.
That means that the $4^{th}$ root of $z$ is $~r\left(\cos{\frac{\theta}{4}} + i\sin{\frac{\theta}{4}} \right)~$ and $~r\left(\cos{(\frac{\theta}{4} + 2\pi)} + i\sin{(\frac{\theta}{4} + 2\pi)} \right)~$ and $~r\left(\cos{(\frac{\theta}{4} + 4\pi)} + i\sin{(\frac{\theta}{4}} + 4\pi)\right)~$, etc.
It's true that you can add multiples of $2\pi$ to infinity, so there should be infinite roots of unity. The reason we only get four, in this case, is because—after adding enough multiples of $2\pi$—the values for $\sqrt[4]{z}$ start repeating themselves.
This is why treating the $4^{th}$ root of $i$ like a regular root seems to come out strangely. You were following all the mathematics that you know, but your proof actually stumbled onto something pretty in-depth! 

$^{[1]}~~$This isn't rigorous, but a quick explanation: You can think of the different parts of a complex number, $~z = x + iy~$, as coordinates. It acts just like the regular $(x,y)$ system we usually use. Here, the $x$ of the complex number is the x-component and the $y$ is the y-component. So, $~4 + i7~$ is $(4,7)$ on the complex coordinate system. After doing this, we can perform geometric operations on our complex number much like we do on a regular coordinate system. Eventually we can derive the Euler's Formula stated above.
By the way, when the coordinates of a complex number $re^{i\theta}$ are mapped out onto the complex plane, the "repeated roots" that one gets all lie on the circumference of a circle whose radius is $r$. For the $4^{th}$ root, one gets four points. Continuing to add multiples of $2\pi$ will just repeat these four points over and over.
