How many values does $\sqrt{\sqrt{i}}$ have? Wolfram says, there are only two roots, but $\sqrt{i}$ already gives two roots. So if we express them in Cartesian form we can take square roots of them separately and end up with four roots.
$$\sqrt{i}=e^{\frac{i\pi}{4}}=\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}$$
But also
$$\sqrt{i}=e^{\frac{-3i\pi}{4}}=-\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}$$
Then take square roots of each of those and you end up with 
$$\sqrt{\sqrt{i}}=e^{\frac{i\pi}{8}},e^{\frac{5i\pi}{4}},e^{\frac{-3i\pi}{4}},e^{\frac{-7i\pi}{4}}$$
It works, doesn't it? Raise each of those to the power of 4 and you get $i$. What am I missing here?
 A: Your list of four solutions only has two - as has been pointed out you listed two of them twice. (In the original post, anyway...) But there are four roots.
First, $i=e^{i\pi/2}=e^{i5\pi/2}$ leads gives two values for $\sqrt i$, namely $e^{i\pi/4}$ and $e^{i5\pi/4}$. Each of these has two square roots: Since $e^{i\pi/4}=e^{i9\pi/4}$ it has the two square roots $e^{i\pi/8}$ and $e^{i9\pi/8}$. Similarly $e^{i5\pi/4}=e^{i13\pi/4}$ has the two square roots $e^{i5\pi/8}$ and $e^{i13\pi/8}$. For a total of four values of $\sqrt{\sqrt i}$, namely $$e^{i\pi/8},e^{i5\pi/8},e^{i9\pi/8},e^{i13\pi/8}.$$
NOTE Others have suggested that although a complex number has two square roots, the notation $\sqrt z$ refers to only one of them. I think it's wrong to put it that way; that notation refers to only one of them if we have clearly stated in advance which "branch" we're referring to. But Wolfram is wrong in any case. By my lights, $i$ has two square roots, each of which has two square roots, for a total of four as above. But if we are saying that the notation $\sqrt z$ refers to one square root, then Wolfram should say there's only one value for $\sqrt{\sqrt i}$, namely the one square root of the one square root. 
I think four is the right number. With that convention there's only one; there's no sensible way to give exactly two values, as WA does. You can't believe everything they tell you.
A: Trick is to introduce an unknown, $z$. Write $z=\sqrt{ \sqrt{i}}$. We want to know how many values are possible for $z$. Squaring we get $z^2=\surd i$. Again squaring we  get $z^4-i=0$. This is a polynomial of degree 4 over the complex  numbers. So it has 4 roots. Taking the derivative shows that the roots are distinct.
SO we have 4 distinct values for the number.
EDIT (to clarify the doubts raised by SHailesh and Piquito in the comments):
For a  polynomial $f(z)$ to have $\alpha$ as repeated root means it is of the form $(z-\alpha)^2 g(z)$ for some other polynomial $g(z)$.Calculating derivative of $f(z)$ using product rule makes it clear that $\alpha$ will also be a root of the derivative $f'(z)$; and converse is also true.
@Piquito: Fundamental theorem of algebra states that the number of roots is equal to the degree of the polynomial only when we count them with multiplicity  and include    complex roots, real irrational roots.
A: As has been pointed out in the other answers, there are four complex numbers $z$ with the property that $z^4=i$. If you would like to understand the behavior exhibited by WolframAlpha, however, you should consider the fact that its square root function is (in fact) a function built on top of Mathematica's Sqrt command. It returns exactly one number, namely 
$$\sqrt{r e^{i\theta}} = \sqrt{r} e^{i\theta/2},$$
when $r\geq0$ and $-\pi<\theta\leq\pi.$
As a result, $\sqrt{\sqrt{i}}$ is uniquely determined to be
$$\sqrt{\sqrt{i}} = \cos \left(\frac{\pi }{8}\right)+i \sin \left(\frac{\pi }{8}\right) \approx 0.92388 + 0.382683 i.$$
That is exactly what Wolfram Alpha has told you in the first couple of pods.
When you say that Alpha says there are two roots, I assume that you are referring to the last couple of pods that refer to "all 2nd roots of $(-1)^{1/4}$".  First off, those pods would be considered secondary. They arise because WolframAlpha typically produces secondary pods containing related information in an effort to respond to the various things that a human user might have meant in their input. 
The reason that there are two is because sqrt(sqrt(i)) essentially parses as
sqrt(a complex number)

so you get the two roots of that complex number in the secondary pod. You would get the same result, if you type sqrt((1+i)/sqrt(2)) because 
$$\sqrt{i} = (1+i)/\sqrt{2}.$$
Finally, if you want all four fourth roots of $i$, simply type
all fourth roots of i

You can get more information on how WolframAlpha deals with roots in this blog post.
A: We can write $i$ as $i=e^{i(\pi/2+2k\pi)}$ for any integer $k$.  Then, taking a square root, we see that 
$$\sqrt{i}=e^{i(\pi/4+k\pi)} \tag 1$$
for any integer $k$.  Note that there are two distinct values of $\sqrt i$ in $(1)$; for $k=-1,0$, we have $\sqrt i = e^{-3i\pi/4}$ and $\sqrt i = e^{i\pi/4}, respectively. 
Taking the square root of the right-hand side of $(1)$ gives
$$\sqrt{\sqrt{i}}=e^{i(\pi/8+k\pi/2)}$$
for which there are $4$ distinct values corresponding to $k=-2,-1,0,1$.  The solutions are $e^{-i7\pi/8}$, $ e^{-i3\pi/8}$, $e^{i\pi/8}$, and $e^{i5\pi/8}$
