Square each side and solve to get y=0 or y=i.

I would discard y=0 since we have no indication as to where a zero popped up in the original infinite radical.

But lets take another approach.  Express each sqrt as a power of 1/2.

Then y = (i*(i*(i*(....)^1/2)^1/2)^1/2.

Again playing a bit loose with exponentiation (not well-defined for infinite products), lets distribute all the powers.  Then that produces y=i^1/2*i^1/4*i^1/8*.....

(At this point I'll introduce another simplifying convention: the principal roots of i are the roots of i that are closest to the positive real axis from above.)

Everything has the same base, so lets collect all the exponents into a common power.  Then y=i^(1/2+1/4+1/8+....)

Now the exponent is a well-known geometric series with a value of 1. Then y=i is our answer.

I suspect this approach, resulting in y=i, is what Ady is intending, based on the D1 rating. But what if we chose a different system to choose the roots?  Lets take i to its trigonometric form: i = cos (pi/2) + i*sin(pi/2).  

If we were to choose some finite number of roots other than the principal root we will get something like y = cos (pi/2 + r/2^n) + i*sin(pi/2 + r/2^n), where n is the deepest root and r is some integer between 0 and 2^n.  An example, taking non-principal roots of i^1/2 and i^1/8 may give us y = cos (pi/2 + pi*5/8) + i*sin(pi/2 + pi*5/8) = -(sqrt[2+sqrt(2)])/2 - i*(sqrt[2-sqrt(2)])/2.

Things get worse as n approaches infinity as then we can get arbitrarily close to any angle we want, thus making the potential domain of y being the entire circumference of the unit circle centered at the origin of the complex plane.

In summary, if we take "sensible" simplifying restrictions then y=i is the solution but there's plenty of complications that leave this problem as ill-defined.

I'm not saying this is a bad problem; actually it's really good for doing a deep dive into what chaos root extraction can do in the complex numbers if you aren't careful.  Especially with the infinite radicals/product mixed in, which have their own complications.