When you have something like "infinity," you have to realize that it's
not a number. Usually what you mean is some kind of limiting process.
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. Cool!
Is it really true? Try plugging in a big value of x. Or recognize this
limit as a variation of the definition of e. Either way, it's true. 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.
Does that help any?