Calculus Limits using L'Hopitals Rule

I asked this question earlier, and I got a good answer, but I didn't get any specifics. Can I get a detailed step by step guide on how to solve this problem? I'll post my work and where I'm confused.
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Problem:

Find the following limit using L'Hopitals Rule:

lim X^X
x->0+
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The Earlier Answer:

"Exponential indeterminant forms must be changed to fractions before L'Hopital may be applied. This is done by finding the limit of Ln(f(x)) rather than f(x) itself,

If L = lim X^X then
ln(L) = Lim x ln(X) = Lim ln(X) / x^-1 approaches - ∞/+∞ is negative and indeterminant. Applying :L'Hopital ln(L) = Lim x^-1 / -x^-2 = -x which approaches 0. Thus L = e^0 = 1."
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Current Confusion:

I now understand that you have to change exponential indeterminate forms into fractions in order to use L'Hopital's Rule, and that you use the natural log to do so.

** For the sake of typing/cleanliness: The limit of x -> 0+ is going to be written as "lim(x->0+)", and limit (answer) is going to be written as "L".

So,

L = lim(x->0+) x^x
=> (ln)L = lim(x->0+) (ln)x^x
=> (ln)L = lim(x->0+) xlnx
=> (ln)L = lim(x->0+) lnx/x^-1
L'Hopitals Rule Applied:
=> (ln)L = lim(x->0+) (1/x)/(-1/x^2)
=> (ln)L = lim(x->0+) (1/x)*(x^2/-1)
=> (ln)L = lim(x->0+) x^2/-x
=> (ln)L = lim(x->0+) -x
=> (ln)L = -(0)
=> (ln)L = 0

I can do this part with no issues. My confusion is with the second half of the problem:

=> L = lim(x->0+) x^x
=> L = lim(x->0+) e^lnx^x
=> L = e^lim(x->0+) lnx^x
=> L = e^0
=> L = 1

I just don't know where "e" came from. Does "e" being raised to "ln___" cause the "e" and "ln" to cancel each other out?
I understand that when "e" is raised to "lim(x->0+) lnx^x", that you can just plug in 0 because we solved for (ln)L in part 1. I just don't get where "e" came from.

No, no, no!!!!!

You calculated the limit of ln L. The limit of L is L = e^(ln L) = e^0