The first reason that comes to mind is that if you think of $i=\sqrt{-1}$, then you may be tempted to argue that
$\begin{align*} i^2&=\sqrt{-1}\sqrt{-1}\\ &=\sqrt{(-1)(-1)}\\ &=\sqrt{1}\\ &=1, \end{align*}$ which is clearly false, as $i^2=-1$. I've always thought of $i$ as something you square to get $-1$, rather than thinking of it as a square root.
There is a bit more complicated, but more thorough explanation, however, involving complex analysis. The problem lies in trying to take fractional exponents of negative numbers, e.g. $(-1)^{1/2}$. In high school you are told that square roots and logarithms are not defined for negative numbers, but the truth is that they do exist, but only as complex numbers.
For a complex number $z$ (which would include all negative real numbers) we define the natural logarithm of $z$ as $\mathcal L_\tau(z)=\ln|z|+i\;\arg_\tau(z),$ where $\ln{|z|}$ is the real natural logarithm of the modulus of $z$ (which is always positive and real), and $\arg_\tau(z)$ is the argument of the complex number $z$ (the angle it makes with the positive real axis in the complex plane) with values in the range $[\tau,\tau+2\pi)$. We then have that $\begin{align*} \exp{(\mathcal L_\tau(z))}&=\exp(\ln|z|+i\;\arg_\tau(z))\\ &=e^{\ln|z|}e^{i\;\arg_\tau(z)}\\ &=|z|(\cos(\arg_\tau(z))+i\;\sin(\arg_\tau(z)))\\ &=r(\cos(\theta)+i\;\sin(\theta))\\ &=z, \end{align*}$ regardless of our choice of of $\tau$.
Therefore, when can use this definition to define $z^{\alpha}$, where $\alpha$ is any complex number. We simply define it as follows: $z^{\alpha}:=\exp(\alpha \mathcal L_\tau(z)),$ for some choice of $\tau$. This makes sense since $\mathcal L_\tau(z^n)=n\mathcal L_\tau(z)$, for $n$ an integer, and since $\exp$ is the inverse of $\mathcal L_\tau$.
However, since the choice of $\tau$ is not unique, then we can have different values of $z^{\alpha}$ for different choices of $\tau$. This makes complex (and fractional) powers of complex (and negative) numbers trickier than we'd like.