You can't just do a square root in reverse, because as 
Yousername pointed out, there can be a positive or a negative component to it. Take for example the square root of 4. While you could say that the square root of 4 is 2, and be partially correct, the more correct answer is "Positive or negative 2". Why is this? Well, lets look at each case individually:
2² = 4
This first case is pretty simple; We know that 2² is equal to 2*2, which is equal to 2+2, which is equal to 4.
-2² = 4
This second case is the "tricky" one. We know that this expression is true, but why is it true? Well, we can factor it out to be $(-1*2)^{2} = 4$. Distribution of the power allows for $-1^{2}*2^{2} = 4$. We proved in the simple case that $2^{2} = 4$, so we can substitute that in here: $-1^{2}*4 = 4$. Finally, solving for $-1^{2}$ we can divide both sides by 4 and find that $-1^{2} = 1$. Now you're probably wondering - "Wait, what does any of this have to do with imaginary numbers?"
Because we have proven that $-1^{2} = 1$, we can prove that the square of any negative number, n, will be equal to the square of the positive n. This can be done by proving that any negative number, n, can be factored into $-1 * n$, which of course is simply the definition of a negative number. Therefore, it is impossible to take the square root of any negative number and have a real value, since squares can only be positive. The way around this was to simply invent imaginary numbers. If we follow the same logic that a negative number can be factored, it can be seen thus: $√(-n) = √(-1*n) = √(-1)*√(n)_{+}$. Note here that we are only interested in the positive square root of n, not the negative.
But, we still have the issue of $√(-1)$. We can't use our factoring trick on it, because we would simply end up with $1*√(-1)$ again. This is why we need to define i as the square root of negative one. Knowing this, calculating the powers of i becomes very easy. $i^{1} = i, i^{2} = -1, i^{3} = -1*i, ...$. If we look at $i^{4}$, we can break it into the component parts $i^{2}*i^{2}$, since we know that exponents with the same base can be added. Applying substitution, $√(-1)^{2}*√(-1)^{2} = -1 * -1 = 1$ so $i^{4} = 1$.
Therefore, taking $i^{n}$, for any odd , the result will be either i or -i. For any even n, the result will be either 1 or -1.
Going back to your original question on if there was a flaw in your logic. Because the 4th root of a number is equivalent to the square root of a square root, and we have proven that the true square root of a number can be positive or negative, Yousername is correct in saying that the fourth root can be positive or negative 1.
I know that was long winded and probably boring, but I was bored and had nothing else to do
(account deleted) 11 Nov 2020 16:05