Discussion on IIT-JEE

 infinity is not really a number it is a concept. any number other than one, raised to an infinite power, would be infinite if the base is greater than 1, and zero if the base were less than 1  So if you have "1^infinity" what you really have is some kind of limit:
the base isn't really 1, but is getting closer and closer to 1 perhaps
while the exponent is getting bigger and bigger, like maybe (x+1)^(1/x) as x->0+.

The question is, which is happening faster, the base getting close to
1 or the exponent getting big? To find out, let's call:

L = lim x->0 of (x+1)^(1/x)

Then:

ln L = lim x->0 of (1/x) ln (x+1) = lim x->0 of ln(x+1) / x

So what's that? As x->0 it's of 0/0 form, so take the derivative of the
top and bottom. Then we get lim x->0 of 1/(x+1) / 1, which = 1.
So ln L = 1, and L = e.

The limit is of the 1^infinity form, but in this case it's e, not 1. Try
repeating the work with (2/x) in the exponent, or with (1/x^2), or with
1/(sqrt(x)), and see how that changes the answer.

That's why we call it indeterminate - all those different versions of
the limit approach 1^infinity, but the final answer could be any
number, such as 1, or infinity, or undefined. You need to do more work to determine the answer, so 1^infinity by itself is not determined yet. In other words, 1 is just one of the answers of 1^infinity.