The square root of $1$ [duplicate]

I know this is wrong but I don't know why. In the set of complex numbers: $\sqrt1 = \sqrt{i^2\cdot i^2} = i\cdot i = -1$ What is wrong with this?

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marked as duplicate by MJD, Pedro Tamaroff♦, Zev Chonoles, Git Gud, Maisam HedyellooJul 28 '13 at 19:54

What are you using $\sqrt{}$ to mean? – Chris Eagle Jul 28 '13 at 19:40
See this post, please. The square root function is not well-defined on the complex numbers. – amWhy Jul 28 '13 at 19:41
Note that $\sqrt{xy}=\sqrt{x}\sqrt{y}$ is no longer an identity for complex numbers. – Adriano Jul 28 '13 at 19:43

One must take care when defining $\sqrt{}$ for complex numbers; it's not generally true that the square root of a product is the product of the square roots.
Last time I've checked $-1$ squared was equal to one. That would be the usual way of checking if the root is calculated correctly.
$\sqrt x$ is normally defined to be the non-negative number $y$ such that $y^2 = x$. – MJD Jul 28 '13 at 19:56