Is there any value possible for i^i

As we all know about the topic of complex numbers, we are familiar with the term iota (i), where i = √(-1). A question arises that is there any value possible for iⁱ.
So, its simple answer is yes, there is a value for iⁱ.  The below solution is mentioned for it.
We have to find the value of iⁱ. So, Let y = iⁱ
Taking ln on both sides,

ln(y)= i ln(i) ----- ( i )      [ ln (ab) = b*ln(a) ]

Now, for solving ln(i), we have to understand the following concept :
In the polar representation of complex numbers, we write z = re^iθ, where –

z = a + ib,
r = |a2 + b2|
θ = tan-1(b/a),
So, taking log on both sides of the equation z = reiθ  
ln(z) = ln(r) + iθ             [ln(ea) = a, and ln(a*b) = ln(a) + ln(b)]
Putting the value of z, r and θ in the above equation
ln(a+ib) = ln(|a2 + b2|) + i*tan-1(b/a)

So, writing ln(i) = ln(0 + 1i), and applying the above formula

ln(0+1i) = ln(|02 + 12|) + i*tan-1(1/0)
ln(i) = ln1 + i*∏/2     [ tan-1(1/0) = tan-1(∞) = ∏/2 ]
ln(i) = i*∏/2           [ ln1 = 0 ]

Now putting the value of ln(i) in equation ( i )

ln(y) = i * ( i*∏/2 )
ln(y) = i2 * ∏/2  
ln(y) = -1 * ∏/2    [i2 = -1]
ln(y) = -∏/2   
y = e -∏/2           [ln(a) = b ⇒ a = eb]

As we have assumed y = iⁱ.
So, 

ii =  e -∏/2    

If we calculate the value for e ^-∏/2 with the help of a calculator, we get its approximate value as 0.20788.

ii = 0.20788